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Calculating molecular weights.

"}, "variablesTest": {"maxRuns": 100, "condition": ""}, "variables": {"h_1": {"description": "", "name": "h_1", "definition": "random(1..12)", "group": "Ungrouped variables", "templateType": "anything"}, "ca": {"description": "

Standard atomic weight of calcium

", "name": "ca", "definition": "40.078", "group": "Ungrouped variables", "templateType": "anything"}, "o_1": {"description": "", "name": "o_1", "definition": "random(1..12)", "group": "Ungrouped variables", "templateType": "anything"}, "Molecule_weights": {"description": "", "name": "Molecule_weights", "definition": "[ 19, 12 ]", "group": "Ungrouped variables", "templateType": "list of numbers"}, "h": {"description": "

Standard atomic weight of hydrogen

", "name": "h", "definition": "1.008", "group": "Ungrouped variables", "templateType": "anything"}, "c": {"description": "

Standard atomic weight of carbon

", "name": "c", "definition": "12.011", "group": "Ungrouped variables", "templateType": "anything"}, "o": {"description": "

Standard atomic weight of oxygen

", "name": "o", "definition": "15.999", "group": "Ungrouped variables", "templateType": "anything"}, "Molecule_indices": {"description": "", "name": "Molecule_indices", "definition": "deal(2)", "group": "Ungrouped variables", "templateType": "anything"}, "c_1": {"description": "", "name": "c_1", "definition": "random(1..12)", "group": "Ungrouped variables", "templateType": "anything"}, "p": {"description": "

Standard atomic weight of phosporus

", "name": "p", "definition": "30.974", "group": "Ungrouped variables", "templateType": "anything"}, "Molecule_names": {"description": "", "name": "Molecule_names", "definition": "[ \"H_2O\", \"CO_2\" ]", "group": "Ungrouped variables", "templateType": "list of strings"}}, "preamble": {"js": "", "css": ""}, "variable_groups": [], "statement": "

Calculate the molecular weights of the following molecules. Use the standard atomic weights given below:

Calcium ($Ca$): $\\var{ca}$

Carbon ($C$): $\\var{c}$

Hydrogen ($H$): $\\var{h}$

Oxygen ($O$): $\\var{o}$

Phosporus ($P$): $\\var{p}$

Please enter your answers as decimals, not as fractions. Give your answers to 3 decimal places.

If you would like to see how to do this question, click on 'Reveal answers' at the bottom of the page.

", "tags": ["molecular mass", "Molecular mass", "molecular weight", "Molecular weight", "Relative molecular mass", "RMM"], "ungrouped_variables": ["h_1", "ca", "o_1", "p", "Molecule_weights", "Molecule_indices", "c_1", "h", "o", "c", "Molecule_names"], "parts": [{"correctAnswerStyle": "plain", "showFeedbackIcon": true, "prompt": "

Water - $H_2O$.

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Methane - $CH_4$.

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Glucose - $C_6H_{12}O_6$.

", "variableReplacements": [], "showCorrectAnswer": true, "mustBeReduced": false, "maxValue": "6*c+12*h+6*o+0.0005", "allowFractions": false, "type": "numberentry", "correctAnswerFraction": false, "minValue": "6*c+12*h+6*o-0.0005", "mustBeReducedPC": 0, "notationStyles": ["plain", "en", "si-en"], "scripts": {}, "variableReplacementStrategy": "originalfirst", "marks": 1}, {"correctAnswerStyle": "plain", "showFeedbackIcon": true, "prompt": "

Calcium Phosphate - $Ca_3(PO_4)_2$.

", "variableReplacements": [], "showCorrectAnswer": true, "mustBeReduced": false, "maxValue": "3*ca+2*p+8*o+0.0005", "allowFractions": false, "type": "numberentry", "correctAnswerFraction": false, "minValue": "3*ca+2*p+8*o-0.0005", "mustBeReducedPC": 0, "notationStyles": ["plain", "en", "si-en"], "scripts": {}, "variableReplacementStrategy": "originalfirst", "marks": 1}], "advice": "

a) Water - $H_2O$.

A water molecule has 2 hydrogen atoms ($H$) and 1 oxygen atom ($O$) so the molecular weight is

$2 \\times \\var{h} + \\var{o} = \\var{2 * h + o}$.

b) Methane - $CH_4$.

A methane molecule has 1 carbon atom ($C$) and 4 hydrogen atoms ($H$) so the molecular weight is

$\\var{c} + 4 \\times \\var{h} = \\var{c + 4 * h}$.

c) Glucose - $C_6H_{12}O_6$.

A glucose molecule has 6 carbon atoms ($C$), 12 hydrogen atoms ($H$) and 6 oxygen atoms ($O$) so the molecular weight is

$6 \\times \\var{c} + 12 \\times \\var{h} + 6 \\times \\var{o} = \\var{6 * c + 12 * h + 6 * o}$.

d) Calcium phospate - $Ca_3(PO_4)_2$.

A calcium phosphate molecule has 3 calcium atoms ($Ca$) and 2 phosphate ions ($PO_4$). We don't need to worry too much about what an ion is here, we just need to know that $(PO_4)_2$ means there are 2 lots of $PO_4$ which means a total of 2 phosporus atoms ($P$) and $4 \\times 2 = 8$ oxygen atoms ($O$). Therefore, the molecular weight is

$3 \\times \\var{ca} + 2 \\times \\var{p} + 8 \\times \\var{o} = \\var{3 * ca + 2 * p + 8 * o}$.

", "rulesets": {}, "type": "question"}, {"name": "Moles of a substance 1", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Jamie Antoun", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/266/"}], "functions": {}, "variablesTest": {"maxRuns": 100, "condition": ""}, "rulesets": {}, "preamble": {"js": "", "css": ""}, "variable_groups": [], "statement": "

Answer the following questions. Please enter your answers as decimals, not as fractions. Give your answers to 2 decimal places.

If you would like to see how to do this question, click on 'Reveal answers' at the bottom of the page.

", "tags": ["molar", "molar calculations", "mole", "moles"], "ungrouped_variables": ["b", "a", "glucose", "NaCl"], "parts": [{"type": "gapfill", "showFeedbackIcon": true, "prompt": "

Glucose has a molecular weight of $\\var{glucose}$. How much does $\\var{0.25 * a}$ mole(s) of glucose weigh in grams?

[[0]] grams

", "variableReplacements": [], "showCorrectAnswer": true, "gaps": [{"correctAnswerStyle": "plain", "showFeedbackIcon": true, "variableReplacements": [], "type": "numberentry", "mustBeReduced": false, "variableReplacementStrategy": "originalfirst", "allowFractions": false, "showCorrectAnswer": true, "correctAnswerFraction": false, "minValue": "0.25 * a * glucose - 0.005", "mustBeReducedPC": 0, "notationStyles": ["plain", "en", "si-en"], "maxValue": "0.25 * a * glucose + 0.005", "scripts": {}, "marks": 1}], "marks": 0, "scripts": {}, "variableReplacementStrategy": "originalfirst"}, {"type": "gapfill", "showFeedbackIcon": true, "prompt": "

Sodium chloride has a molecular weight of $\\var{NaCl}$. How much does $\\var{0.25 * b}$ mole(s) of sodium chloride weigh in grams?

[[0]] grams

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To answer these questions we use the formula

$\\text{molecular weight} \\times \\text{number of moles} = \\text{mass of substance (in grams)}$.

Glucose has a molecular weight of $\\var{glucose}$. How much does $\\var{0.25 * a}$ mole(s) of glucose weigh in grams?

Solution:

Putting our numbers into the formula we find that $\\var{0.25 * a}$ mole(s) of glucose weighs

$\\var{glucose} \\times \\var{0.25 * a} = \\var{glucose * 0.25 * a}$ grams.

Sodium chloride has a molecular weight of $\\var{NaCl}$. How much does $\\var{0.25 * b}$ mole(s) of sodium chloride weigh in grams?

Solution:

Putting our numbers into the formula we find that $\\var{0.25 * b}$ mole(s) of sodium chloride weighs

$\\var{NaCl} \\times \\var{0.25 * b} = \\var{NaCl * 0.25 * b}$ grams.

", "variables": {"NaCl": {"description": "

Molecular weight of sodium chloride.

", "name": "NaCl", "definition": "58.44", "group": "Ungrouped variables", "templateType": "anything"}, "a": {"description": "", "name": "a", "definition": "random(1..10)", "group": "Ungrouped variables", "templateType": "anything"}, "b": {"description": "", "name": "b", "definition": "random(1..10 except a)", "group": "Ungrouped variables", "templateType": "anything"}, "glucose": {"description": "

molecular weight of glucose

", "name": "glucose", "definition": "180.16", "group": "Ungrouped variables", "templateType": "anything"}}, "metadata": {"licence": "Creative Commons Attribution 4.0 International", "description": "

Calculating mass of a substance given molecular weight and number of moles.

"}, "type": "question"}, {"name": "Moles of a substance 2", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Jamie Antoun", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/266/"}], "tags": ["mole", "molecular mass", "Molecular mass", "molecular weight", "Molecular weight", "moles", "RMM"], "metadata": {"description": "

Calculate the number of moles in a given mass of a substance given the molecular weight.

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

Answer the following questions. Please enter your answers as decimals, not as fractions. Give your answers to 2 decimal places.

If you would like to see how to do this question, click on 'Reveal answers' at the bottom of the page.

", "variable_groups": [], "advice": "

To answer these questions we use the formula

$\\dfrac{\\text{mass of substance (in grams)}}{\\text{molecular weight}} = \\text{number of moles}$.

Glucose has a molecular weight of $\\var{glucose}$. How many moles of glucose are there in $\\var{5 * a}$ grams?

Solution:

Putting our numbers into the formula we find that there are

$\\begin{align} \\dfrac{\\var{5 * a}}{\\var{glucose}} &= \\var{(5 * a) / glucose} \\text{ moles}\\\\ &= \\var{precround(((5 * a) / glucose), 2)} \\text{ moles to 2 d.p.} \\end{align}$

in $\\var{5 * a}$ grams of glucose.

Sodium chloride has a molecular weight of $\\var{NaCl}$. How many moles of sodium chloride are there in $\\var{5 * b}$ grams?

Solution:

Putting our numbers into the formula we find that there are

$\\begin{align}\\dfrac{\\var{5 * b}}{\\var{NaCl}} &= \\var{(5 * b) / NaCl} \\text{ moles} \\\\ & = \\var{precround(((5 * b) / NaCl), 2)} \\text{ moles to 2 d.p.} \\end{align}$.

in $\\var{5 * b}$ grams of sodium chloride.

", "functions": {}, "rulesets": {}, "variablesTest": {"condition": "", "maxRuns": 100}, "parts": [{"prompt": "

Glucose has a molecular weight of $\\var{glucose}$. How many moles of glucose are there in $\\var{5 * a}$ grams?

[[0]] moles

Sodium chloride has a molecular weight of $\\var{NaCl}$. How many moles of sodium chloride are there in $\\var{5 * b}$ grams?

[[0]] moles

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molecular weight of glucose

", "templateType": "anything", "name": "glucose", "definition": "180.16"}, "NaCl": {"group": "Ungrouped variables", "description": "

molecular weight of sodium chloride

", "templateType": "anything", "name": "NaCl", "definition": "58.44"}}, "ungrouped_variables": ["a", "NaCl", "b", "glucose"], "preamble": {"css": "", "js": ""}, "type": "question"}, {"name": "Solutions 1", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Jamie Antoun", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/266/"}], "statement": "

Answer the following questions. Please enter your answers as decimals, not as fractions. Enter your answers to 2 decimal places.

If you would like to see how to do this question, click on 'Reveal answers' at the bottom of the page.

", "ungrouped_variables": ["a", "c", "b", "d"], "variable_groups": [], "variablesTest": {"maxRuns": "100", "condition": ""}, "functions": {}, "advice": "

To answer these questions we use the formula

$\\dfrac{\\text{number of moles of substance}}{\\text{volume of liquid (in litres)}} = \\text{concentration (in mol/L)}$.

a)

$\\var{a}$ mole(s) of a substance are dissolved in $\\var{b}$ litre(s) of a liquid to make a solution. What is the concentration of the solution in M (mol/L)?

Solution:

Putting our numbers into the formula, we find that the concentration is

$\\begin{align}\\dfrac{\\var{a}}{\\var{b}} & = \\var{a / b} M \\\\ & = \\var{precround((a / b), 2)} M \\text{ to 2 d.p.} \\end{align}$

b)

$\\var{0.25 * c}$ mole(s) of a substance are dissolved in $\\var{250 * d}$ml of a liquid to make a solution. What is the concentration of the solution in M (mol/L)?

Solution:

The formula uses the volume of liquid in litres so we first have to convert $\\var{250 * d}$ml to a volume in litres. There are $1000$ml in 1L so $\\var{250 * d}$ml is equal to

$\\dfrac{\\var{250 * d}}{1000} = \\var{250 * d / 1000}L$.

Putting our numbers into the formula, we find that the concentration is

$\\begin{align}\\dfrac{\\var{0.25 * c}}{\\var{250 * d / 1000}} & = \\var{(0.25 * c) / (250 * d / 1000)} M \\\\ & = \\var{precround(((0.25 * c) / (250 * d / 1000)), 2)} M \\text{ to 2 d.p.} \\end{align}$

", "metadata": {"licence": "Creative Commons Attribution 4.0 International", "description": "

Practice calculating concentration of solutions

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$\\var{a}$ mole(s) of a substance are dissolved in $\\var{b}$ litre(s) of a liquid to make a solution. What is the concentration of the solution in M (mol/L)?

[[0]] M

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$\\var{0.25 * c}$ mole(s) of a substance are dissolved in $\\var{250 * d}$ml of a liquid to make a solution. What is the concentration of the solution in M (mol/L)?

[[0]] M

", "variableReplacements": [], "marks": 0}], "type": "question"}, {"name": "Solutions 2", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Jamie Antoun", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/266/"}], "variablesTest": {"maxRuns": 100, "condition": ""}, "variable_groups": [], "rulesets": {}, "preamble": {"css": "", "js": ""}, "metadata": {"licence": "Creative Commons Attribution 4.0 International", "description": "

Practice calculating number of moles of a substance given the concentration and volume of a solution.

"}, "tags": ["concentration", "molarity", "moles", "solutions"], "ungrouped_variables": ["a", "c", "b", "d"], "parts": [{"type": "gapfill", "scripts": {}, "showFeedbackIcon": true, "prompt": "

How many moles of glucose are there in $\\var{a}$L of a $\\var{0.25 * b}$M (mol/L) solution?

[[0]] moles

How many moles of glucose are there in $\\var{25 * c}$mL of a $\\var{0.25 * d}$M (mol/L) solution?

[[0]] moles

", "gaps": [{"type": "numberentry", "allowFractions": false, "showPrecisionHint": false, "showFeedbackIcon": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerFraction": false, "precisionMessage": "You have not given your answer to the correct precision.", "mustBeReduced": false, "variableReplacements": [], "maxValue": "(25 * c / 1000) * 0.25 * d + 0.005", "marks": 1, "precisionType": "dp", "precisionPartialCredit": 0, "correctAnswerStyle": "plain", "scripts": {}, "strictPrecision": false, "showCorrectAnswer": true, "minValue": "(25 * c / 1000) * 0.25 * d - 0.005", "variableReplacementStrategy": "originalfirst", "mustBeReducedPC": 0, "precision": "2"}], "showCorrectAnswer": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "marks": 0}], "variables": {"d": {"name": "d", "description": "", "definition": "random(4..20)", "group": "Ungrouped variables", "templateType": "anything"}, "b": {"name": "b", "description": "", "definition": "random(1..10)", "group": "Ungrouped variables", "templateType": "anything"}, "a": {"name": "a", "description": "", "definition": "random(1..5)", "group": "Ungrouped variables", "templateType": "anything"}, "c": {"name": "c", "description": "", "definition": "random(2..10)", "group": "Ungrouped variables", "templateType": "anything"}}, "advice": "

To answer these questions, we use the formula

$\\text{volume of liquid (in litres)} \\times \\text{concentration (in mol/L)} = \\text{number of moles of substance}$.

a)

How many moles of glucose are there in $\\var{a}$L of a $\\var{0.25 * b}$M (mol/L) solution?

Solution:

Putting our numbers into the formula, we find that there are

$\\var{a} \\times \\var{0.25 * b} = \\var{a * 0.25 * b}$ moles

of glucose in $\\var{a}$L of a $\\var{0.25 * b}$M (mol/L) solution.

b)

How many moles of glucose are there in $\\var{25 * c}$mL of a $\\var{0.25 * d}$M (mol/L) solution?

Solution:

Our formula uses the volume of liquid in litres so first we have to convert $\\var{25 * c}$mL to a volume in litres. There are 1000ml in 1L so $\\var{25 * c}$mL is equal to

$\\dfrac{\\var{25 * c}}{1000} = \\var{25 * c / 1000}$L.

Putting our numbers into the formula, we find that there are

$\\begin{align}\\var{25 * c / 1000} \\times \\var{0.25 * d} & = \\var{(25 * c / 1000) * 0.25 * d} \\text{ moles} \\\\ & = \\var{precround(((25 * c / 1000) * 0.25 * d),2 )} \\text{ moles to 2 d.p.}\\end{align}$

of glucose in $\\var{25 * c}$mL of a $\\var{0.25 * d}$M (mol/L) solution.

", "functions": {}, "statement": "

Answer the following questions. Please enter your answers as decimals, not as fractions. Give your answers to 2 decimal places.

If you would like to see how to do this question, click on 'Reveal answers' at the bottom of the page.

", "type": "question"}, {"name": "Solutions 3", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Jamie Antoun", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/266/"}], "functions": {}, "rulesets": {}, "variablesTest": {"maxRuns": 100, "condition": ""}, "preamble": {"js": "", "css": ""}, "variable_groups": [], "statement": "

Answer the following question. Please enter your answer as a decimal, not as a fraction. Give your answer to 2 decimal places.

Clicking on 'Show steps' will provide you with some prompts to break down the question into smaller parts.

If you would like to see how to do this question, click on 'Reveal answers' at the bottom of the page.

", "tags": ["concentration", "molarity", "moles", "solution", "solutions"], "ungrouped_variables": ["a", "b", "glucose"], "parts": [{"stepsPenalty": 0, "steps": [{"correctAnswerStyle": "plain", "showFeedbackIcon": true, "prompt": "

1) First calculate how many litres there are in $\\var{100 * a}$ml.

", "variableReplacements": [], "type": "numberentry", "mustBeReduced": false, "maxValue": "100 * a / 1000", "allowFractions": false, "showCorrectAnswer": true, "correctAnswerFraction": false, "minValue": "100 * a / 1000", "mustBeReducedPC": 0, "notationStyles": ["plain", "en", "si-en"], "scripts": {}, "variableReplacementStrategy": "originalfirst", "marks": 1}, {"correctAnswerStyle": "plain", "showFeedbackIcon": true, "prompt": "

2) Using the volume in litres you calculated in step 1, work out how many moles of glucose you would need to make a $\\var{0.5 * b}$M solution.

", "variableReplacements": [], "type": "numberentry", "mustBeReduced": false, "maxValue": "(100 * a / 1000) * (0.5 * b)", "allowFractions": false, "showCorrectAnswer": true, "correctAnswerFraction": false, "minValue": "(100 * a / 1000) * (0.5 * b)", "mustBeReducedPC": 0, "notationStyles": ["plain", "en", "si-en"], "scripts": {}, "variableReplacementStrategy": "originalfirst", "marks": 1}, {"correctAnswerStyle": "plain", "showFeedbackIcon": true, "prompt": "

3) Using the number of moles of glucose required which you calculated in step 2, work out the mass of glucose needed using the molecular weight ($\\var{glucose}$).

", "variableReplacements": [], "type": "numberentry", "mustBeReduced": false, "maxValue": "(100 * a / 1000) * (0.5 * b) * glucose + 0.005", "allowFractions": false, "showCorrectAnswer": true, "correctAnswerFraction": false, "minValue": "(100 * a / 1000) * (0.5 * b) * glucose - 0.005", "mustBeReducedPC": 0, "notationStyles": ["plain", "en", "si-en"], "scripts": {}, "variableReplacementStrategy": "originalfirst", "marks": 1}], "showFeedbackIcon": true, "prompt": "

Glucose ($C_6H_{12}O_6$) has a molecular weight of $\\var{glucose}$, what mass of glucose would you need to dissolve in $\\var{100 * a}$ml of water to get a $\\var{0.5 * b}$M solution?

[[0]] grams

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We can break this question up into parts. First we need to know how many moles of glucose we need to make a $\\var{0.5 * b}$M solution using $\\var{100 * a}$ml of water. The calculate this we use the formula

$\\text{volume of liquid (in litres)} \\times \\text{concentration (in mol/L)} = \\text{number of moles of substance}$.

This formula uses the voume in litres so we have to convert $\\var{100 * a}$ml to a volume in litres. There are 1000ml in 1L so $\\var{100 * a}$ml is equal to

$\\dfrac{\\var{100 * a}}{1000} = \\var{100 * a / 1000}$L.

Putting our numbers into the formula, we find that we need

$\\var{100 * a / 1000} \\times \\var{0.5 * b} = \\var{(100 * a / 1000) * 0.5 * b}$ moles

of glucose. Finally, to work out the mass of glucose we need, we use the formula

$\\text{molecular weight} \\times \\text{number of moles} = \\text{mass of substance (in grams)}$.

We are told that glucose has a molecular wight of $\\var{glucose}$ and we have calculated that we need $\\var{(100 * a / 1000) * 0.5 * b}$ moles of glucose. Putting these numbers into the formula, we find that we need

$\\begin{align}\\var{glucose} \\times \\var{(100 * a / 1000) * 0.5 * b} & = \\var{glucose * ((100 * a / 1000) * 0.5 * b)} \\text{ grams} \\\\ & = \\var{precround((glucose * ((100 * a / 1000) * 0.5 * b)), 2)} \\text{ grams to 2 d.p.}\\end{align}$

of glucose to make a $\\var{0.5 * b}$M solution using $\\var{100 * a}$ml of water.

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Molecular weight of glucose

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Practice at a longer calculation to work out how much glucose is needed to make a solution of a given concentration.

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Answer the following question. Please enter your answer as a decimal, not a fraction. Give your answer to 2 decimal places (only round your final answer, use the exact values on your calculator when working out any intermediate steps).

Clicking on 'Show steps' will provide you with some prompts to break down the question into smaller parts.

If you would like to see how to do this question, click on 'Reveal answers' at the bottom of the page.

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$\\var{50 * a}$ml of a $\\var{0.25 * b}$M solution is diluted by adding another $\\var{50 * c}$ml of liquid, what is the new concentration?

[[0]] M

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1) We started with $\\var{50 * a}$ml of liquid and added another $\\var{50 * c}$ml. Calcluate the new volume of liquid in ml.

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2) Convert $\\var{50 * a}$ml to a volume in litres.

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3) Using the value you calculated in step 2, calculate how many moles of the substance we started with. (Because we want to work out our final answer to 2 decimal places, don't round this answer to 2 decimal places, enter it exactly as it appears in your calculator and use the exact number for the rest of the calculations).

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4) Convert the volume you found in step 1 to a volume in litres.

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5) Using your answers to steps 3 and 4, calculate the new concentration of the solution in mol/L.

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$\\var{50 * a}$ml of a $\\var{0.25 * b}$M solution is diluted by adding another $\\var{50 * c}$ml of liquid, what is the new concentration?

Solution:

We started with $\\var{50 * a}$ml of liquid and added another $\\var{50 * c}$ml so the new volume of liquid is

$\\var{50 * a} + \\var{50 * c} = \\var{50 * (a + c)} \\text{ml.}$

$\\var{50 * a}$ml is equal to

$\\dfrac{\\var{50 * a}}{1000} = \\var{50 * a / 1000}\\text{L}$

so the number of moles we started with is

$\\var{50 * a / 1000} \\times \\var{0.25 * b} = \\var{(50 * a / 1000) * 0.25 * b} \\text{ moles}.$

We haven't added any more of the substance so we still have $\\var{(50 * a / 1000) * 0.25 * b}$ moles but this is now dissolved in $\\var{50 * (a + c)}$ml of liquid. $\\var{50 * (a + c)}$ml is equal to $\\var{50 * (a + c) / 1000}$L so the new concentration is

$\\begin{align} \\dfrac{\\var{(50 * a / 1000) * 0.25 * b}}{\\var{50 * (a + c) / 1000}} & = \\var{((50 * a / 1000) * 0.25 * b) / (50 * (a + c) / 1000)} \\text{M} \\\\ & = \\var{precround((((50 * a / 1000) * 0.25 * b) / (50 * (a + c) / 1000)), 2)} \\text{M to 2 d.p.} \\end{align}$

Challenge:

Can you think of a shorter way of doing this calculation? Don't worry if you can't, just go through the steps one by one.

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Calculating the new concentration of a diluted solution.

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The volume of the \"stock\" solution used for dilution in ml.

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atomic weight of chlorine

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The mass concentration of the \"stock\" solution in g/L.

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The mass of the substance (in grams) dissolved in the \"stock\" solution

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molecular weight of sodium chloride

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atomic weight of sodium

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The volume of liquid in the \"stock\" solution in litres.

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The molarity of the \"stock\" solution

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The volume the sample of \"stock\" solution is diluted to in ml.

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Answer the following question. Please give your answers as decimals, not as fractions. Give your answers to 3 decimal places but only round your calculations at the final step, use the exact values in your calculator for intermediate steps.

Clicking on 'Show steps' will provide you with some prompts to break down the question into smaller parts.

If you would like to see how to do this question, click on 'Reveal answers' at the bottom of the page.

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A \"stock\" solution contained $\\var{stock_mass}$g of sodium chloride (NaCl) in $\\var{stock_volume}$L of solution. $\\var{stock_volume_used}$ml of the \"stock\" solution was diluted to $\\var{final_volume}$ml. What is the concentration of the final solution in g/L and M? [Relative atomic masses (Da): Na $\\var{Na}$; Cl $\\var{Cl}$].

[[0]] g/L

[[1]] M

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1) Our \"stock\" solution contains $\\var{stock_mass}$g of sodium chloride (NaCl) in $\\var{stock_volume}$L of solution. Calculate the concentration of the \"stock\" solution in g/L.

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2) Calculate the molecular weight of sodium chloride (NaCl)

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3) Calculate how many moles of sodium chloride there are in $\\var{stock_mass}$g.

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4) Calculate the concentration of the \"stock\" solution in M (mol/L).

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5) Calculate what $\\var{stock_volume_used}$ml is as a volume in litres.

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6) Calculate how many grams of sodium chloride there are in $\\var{stock_volume_used}$ml of the \"stock\" solution using your answers to parts 1 and 5.

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7) Calculate how many moles of sodium chloride there are in $\\var{stock_volume_used}$ml of the \"stock\" solution using your answers to parts 4 and 5.

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8) Calculate what $\\var{final_volume}$ml is as a volume in litres.

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9) Using your answers to parts 6 and 8, calculate the concentration in g/L of the final solution.

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10) 9) Using your answers to parts 7 and 8, calculate the concentration in M of the final solution.

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We have been asked to do quite a lot of calculations in this example so let's break it up into several smaller parts. To work out the concentrations in g/L and M after dilution, we need to first work out the concentrations in g/L and M of the \"stock\" solution. Let's start by working out the concentration of the \"stock\" solution in g/L.

The \"stock\" solution contains $\\var{stock_mass}$g of sodium chloride dissolved in $\\var{stock_volume}$L of solution. To work out the concentration in g/L we use the formula

$\\dfrac{\\text{mass of substance (in grams)}}{\\text{volume of liquid (in litres)}} = \\text{concentration (in g/L)}.$

Putting in our numbers we find that the concentration of the \"stock\" solution in g/L is

$\\dfrac{\\var{stock_mass}}{\\var{stock_volume}} = \\var{stock_mass / stock_volume} \\text{ g/L}.$

Next, we can work out the concentration of the stock solution in M (mol/L). The formula for this is

$\\dfrac{\\text{number of moles of substance}}{\\text{volume of liquid (in litres)}} = \\text{concentration (in mol/L)}.$

We can see that we need to work out how many moles of sodium chloride (NaCl) there are in $\\var{stock_mass}$g. The formula for this is

$\\dfrac{\\text{mass of substance (in grams)}}{\\text{molecular weight}} = \\text{number of moles}.$

We can see that to do this, we are going to need to work out the molecular weight of sodium chloride (NaCl). We have been told the atomic weights of each of the atoms in sodium chloride (NaCl) so we can do this by adding together the atomic weights of all the atoms to get the molecular weight which is

$1 \\times \\var{Na} + 1 \\times \\var{Cl} = \\var{NaCl} \\text{ Da}.$

We can now use this to work out how many moles of sodium chloride there are in $\\var{stock_mass}$g using the formula above. Putting in the numbers we find that there are

$\\dfrac{\\var{stock_mass}}{\\var{NaCl}} = \\var{stock_mass / NaCl} \\text{ moles}$

of sodium chloride in $\\var{stock_mass}$g. We can now work out the concentration of the \"stock\" solution in M (mol/L). The formula is

$\\dfrac{\\text{number of moles of substance}}{\\text{volume of liquid (in litres)}} = \\text{concentration (in mol/L)}.$

We have $\\var{stock_mass}$g of sodium chloride dissolved in $\\var{stock_volume}$L of liquid and we have calculated that there are $\\var{stock_mass / NaCl}$ moles of sodium chloride in $\\var{stock_mass}$g. Putting these numbers into the formula we find that the concentration of the \"stock\" solution in M is

$\\dfrac{\\var{stock_mass / NaCl}}{\\var{stock_volume}} = \\var{stock_molarity} \\text{ M}.$

Let's recap what we have calculated so far. We have worked out that the concentration of the \"stock\" solution in g/L is

$\\var{stock_mass_concentration} \\text{ g/L}$

and in M is

$\\var{stock_molarity} \\text{ M}.$

Since we are diluting $\\var{stock_volume_used}$ml of the \"stock\" solution, we need to work out how much sodium chloride there is in $\\var{stock_volume_used}$ml of the solution using the concentrations we have calculated. We need to work this amount out in both grams and moles, let's start with grams. If we rearrange the formula

$\\dfrac{\\text{mass of substance (in grams)}}{\\text{volume of liquid (in litres)}} = \\text{concentration (in g/L)}.$

we find that

${\\text{volume of liquid (in litres)}} \\times \\text{concentration (in g/L)} = \\text{mass of substance (in grams)}.$

Now, $\\var{stock_volume_used}$ml is equal to

$\\dfrac{\\var{stock_volume_used}}{1000} = \\var{stock_volume_used / 1000} \\text{ L},$

so in $\\var{stock_volume_used}$ml of \"stock\" solution, there are

$\\var{stock_volume_used / 1000} \\times \\var{stock_mass_concentration} = \\var{(stock_volume_used / 1000) * stock_mass_concentration} \\text{ g}$

of sodium chloride. Similarly, we can work out the number of moles of sodium chloride using the formula

${\\text{volume of liquid (in litres)}} \\times \\text{concentration (in mol/L)} = \\text{number of moles of substance}.$

Putting in our numbers, we find that in $\\var{stock_volume_used}$ml of \"stock\" solution, there are

$\\var{stock_volume_used / 1000} \\times \\var{stock_molarity} = \\var{(stock_volume_used / 1000) * stock_molarity} \\text{ moles}$

of sodium chloride.

Finally, we can work out the concentration of the diluted solution. We know that $\\var{stock_volume_used}$ml of \"stock\" solution is diluted to $\\var{final_volume}$ml so the final volume of liquid is $\\var{final_volume}$ml. We have worked out that in $\\var{stock_volume_used}$ml of stock solution there are $\\var{(stock_volume_used / 1000) * stock_mass_concentration}$g ($\\var{(stock_volume_used / 1000) * stock_molarity}$ moles) of sodium chloride and we have not added any more sodium chloride so in our diluted solution we have $\\var{(stock_volume_used / 1000) * stock_mass_concentration}$g ($\\var{(stock_volume_used / 1000) * stock_molarity}$ moles) of sodium chloride in $\\var{final_volume}$l of liquid. The formula for the concentration in g/L is

$\\dfrac{\\text{mass of substance (in grams)}}{\\text{volume of liquid (in litres)}} = \\text{concentration (in g/L)}.$

$\\var{final_volume}$ml is equal to $\\var{final_volume / 1000}$L and so the concentration of the diluted solution in g/L is

$\\begin{align}\\dfrac{\\var{(stock_volume_used / 1000) * stock_mass_concentration}}{\\var{final_volume / 1000}} & = \\var{final_mass_concentration} \\\\ & = \\var{precround(final_mass_concentration, 3)} \\text{ g/L to 3 d.p.}\\end{align}$

The formula for concentration in M (mol/L) is

$\\dfrac{\\text{number of moles of substance}}{\\text{volume of liquid (in litres)}} = \\text{concentration (in mol/L)}$

so the concentration of the diluted solution in M is

$\\begin{align}\\dfrac{\\var{(stock_volume_used / 1000) * stock_molarity}}{\\var{final_volume / 1000}} & = \\var{final_molarity} \\\\ & = \\var{precround(final_molarity, 3)} \\text{ M to 3 d.p.}\\end{align}$

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Example of a dilution calculation involving mass concentration and molarity calculations.

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Working backwards from the concentration of a diluted solution to find the concentration of \"stock\" solution.

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Moalrity of \"stock\" solution.

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Molarity of diluted solution.

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Volume of solution after dilution.

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Volume of \"stock\" solution which was diluted

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Concentration of diluted solution in mM.

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A solution of glucose was prepared by diluting $\\var{stock_volume_used}$ml of the \"stock\" glucose solution to $\\var{final_volume}$ml. The concentration of glucose in the final solution was $\\var{final_concentration}$mM. What was the concentration of glucose in the \"stock\" solution in M (mol/L)?

Solution:

First, we convert $\\var{final_volume}$ml to a volume in litres. There are 1000ml in 1L so $\\var{final_volume}$ml is equal to

$\\dfrac{\\var{final_volume}}{1000} = \\var{final_volume / 1000} \\text{ L.}$

Next, we can calculate how many mmol of glucose there are in the final solution. The formula for this is

$\\text{volume of liquid (in litres)} \\times \\text{concentration (in mmol/L)} = \\text{number of millimoles of substance}$.

[Note: This is slightly different to the formula on the wiki page for molar calculations. This is because we are currently working with mmol and mM as opposed to mol and M. If you prefer, you can first convert all values in mmol and mM to values in mol and M respectively and use the formulas as given on the wiki page. If you would like to read a little more about converting between units, see the wiki page on dimensions in the animal science section.]

Putting in our numbers we find that there are

$\\var{final_volume / 1000} \\times \\var{final_concentration} = \\var{(final_volume / 1000) * final_concentration} \\text{ mmol}$

of glucose in the final solution.

We also need to convert $\\var{stock_volume_used}$ml to a volume in litres. $\\var{stock_volume_used}$ml is equal to

$\\dfrac{\\var{stock_volume_used}}{1000} = \\var{stock_volume_used / 1000} \\text{ L}.$

We didn't add any more glucose when diluting the \"stock\" solution so the \"stock\" solution also contained $\\var{(final_volume / 1000) * final_concentration}$mmol of glucose in $\\var{stock_volume_used / 1000}$L of solution. We calculate the concentration of the \"stock\" solution in mM (mmol/L) using the formula

$\\dfrac{\\text{number of millimoles of substance}}{\\text{volume of liquid (in litres)}} = \\text{concentration (in mmol/L)}$.

[Again, if you prefer, you can convert the amount of substance in mmol to an amount in mol before doing the calculation and use the formulas as given on the wiki page.]

Putting in our numbers, we find that theconcentration of the \"stock\" stock in mM is

$\\dfrac{\\var{(final_volume / 1000) * final_concentration}}{\\var{stock_volume_used / 1000}} = \\var{stock_molarity * 1000} \\text{ mM}.$

Finally, we have been asked to give the answer in M so we have to convert $\\var{stock_molarity * 1000}$mM to a concentration in M. There are 1000mM in 1M so the concentration of the \"stock\" solution in M is

$\\begin{align}\\dfrac{\\var{stock_molarity * 1000}}{1000} & = \\var{stock_molarity}\\text{ M} \\\\ & = \\var{precround(stock_molarity, 3)} \\text{ M to 3 d.p.} \\end{align}$

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[[0]] M

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1) First, convert $\\var{final_volume}$ml to a volume in litres.

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2) Then calculate how much glucose there is in the final solution in mmol.

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3) The next step is to convert $\\var{stock_volume_used}$ml to a volume in litres.

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4) Calculate the concentration of the \"stock\" solution in mM using your answers to parts 2 and 3.

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5) Finally, convert the concentration you found in mM (mmol/L) in part 4 to a concentration in M (mol/L).

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Answer the following question. Please enter your answer as a decimal, not a fraction. Give your answer to 3 decimal places.

Clicking on 'Show steps' will provide you with some prompts to break down the question into smaller parts.

If you would like to see how to do this question, click on 'Reveal answers' at the bottom of the page.

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Test on molar calculations based on Newcastle university wiki page in the animal science section.

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