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Mini-test on concentration of solutions.

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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.

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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

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

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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

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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.

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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.

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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.

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

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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.

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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 ($\\var{glucose}$).

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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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