// Numbas version: exam_results_page_options {"name": "Biomedical Sciences Diagnostic Test - practice", "metadata": {"description": "

A test of basic concepts to do with SI units and concentrations of solutions.

", "licence": "Creative Commons Attribution 4.0 International"}, "duration": 0, "percentPass": 0, "showQuestionGroupNames": false, "shuffleQuestionGroups": false, "showstudentname": true, "question_groups": [{"name": "Group", "pickingStrategy": "all-ordered", "pickQuestions": 1, "questionNames": ["", "", "", "", "", ""], "variable_overrides": [[], [], [], [], [], []], "questions": [{"name": "Identify SI and Non-SI Units - Biomedical Sciences", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Bill Foster", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/6/"}, {"name": "Chris Graham", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/369/"}], "statement": "\n \n \n

The International System of Units (abbreviated SI from the French language name Systeme international d’unites) is the modern form of the metric system. It is the world’s most widely used system of units, both in everyday commerce and in science.

\n \n \n \n

Note: The litre is not an SI unit, but it is accepted for use with the SI. It is not the standard SI unit of volume, however. A litre is defined as a special name for a cubic decimetre $(1 \\mathrm{L} = 1 \\mathrm{dm}^3)$. $1 \\mathrm{L} = 0.001 \\mathrm{m}^3$ (exactly). The original metric system used litres as a base unit.

\n \n ", "variablesTest": {"condition": "", "maxRuns": 100}, "variables": {}, "tags": [], "variable_groups": [], "parts": [{"extendBaseMarkingAlgorithm": true, "scripts": {}, "type": "1_n_2", "variableReplacementStrategy": "originalfirst", "adaptiveMarkingPenalty": 0, "maxMarks": 0, "choices": ["

$\\mathrm{L}$

", "

$\\mathrm{cm}$

", "

$\\mathrm{ft}$

", "

$\\mathrm{km}$

", "

$\\mathrm{in}$

", "

$\\mathrm{m}$

"], "showFeedbackIcon": true, "prompt": "

Choose the base SI unit for length:

", "shuffleChoices": true, "matrix": [0, 0, 0, 0, 0, 1], "variableReplacements": [], "marks": 0, "displayColumns": 6, "showCorrectAnswer": true, "unitTests": [], "minMarks": 0, "distractors": ["", "The centimetre is not the base SI unit of length.", "The foot is an Imperial unit of measurement.", "The kilometre is not the base SI unit for length.", "The inch is an Imperial unit of measurement.", ""], "customMarkingAlgorithm": "", "displayType": "radiogroup", "customName": "", "useCustomName": false, "showCellAnswerState": true}, {"extendBaseMarkingAlgorithm": true, "scripts": {}, "type": "1_n_2", "variableReplacementStrategy": "originalfirst", "adaptiveMarkingPenalty": 0, "maxMarks": 0, "choices": ["

$\\mathrm{cm}^3$

", "

$\\mathrm{kg}$

", "

$\\mathrm{pint}$

", "

$\\mathrm{m}^3$

", "$\\mathrm{V}$"], "showFeedbackIcon": true, "prompt": "

Choose the base SI unit for volume:

", "shuffleChoices": true, "matrix": [0, 0, 0, 1, 0], "variableReplacements": [], "marks": 0, "displayColumns": 6, "showCorrectAnswer": true, "unitTests": [], "minMarks": 0, "distractors": ["", "", "", "", ""], "customMarkingAlgorithm": "", "displayType": "radiogroup", "customName": "", "useCustomName": false, "showCellAnswerState": true}, {"extendBaseMarkingAlgorithm": true, "scripts": {}, "type": "1_n_2", "variableReplacementStrategy": "originalfirst", "adaptiveMarkingPenalty": 0, "maxMarks": 0, "choices": ["

$\\mathrm{L}$

", "

$\\mathrm{gm}$

", "

$\\mathrm{kg}$

", "

$\\mathrm{molar}$

", "

$\\mathrm{mol}$

"], "showFeedbackIcon": true, "prompt": "

Choose the base SI unit for amount of substance:

", "shuffleChoices": true, "matrix": [0, 0, 0, 0, 1], "variableReplacements": [], "marks": 0, "displayColumns": 6, "showCorrectAnswer": true, "unitTests": [], "minMarks": 0, "distractors": ["", "", "", "", ""], "customMarkingAlgorithm": "", "displayType": "radiogroup", "customName": "", "useCustomName": false, "showCellAnswerState": true}, {"extendBaseMarkingAlgorithm": true, "scripts": {}, "type": "1_n_2", "variableReplacementStrategy": "originalfirst", "adaptiveMarkingPenalty": 0, "maxMarks": 0, "choices": ["

$\\mathrm{parsec}$

", "

$\\mathrm{hour}$

", "

$\\mathrm{min}$

", "

$\\mathrm{sec}$

", "

$\\mathrm{t}$

", "

$\\mathrm{s}$

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Choose the base SI unit for time:

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$\\mathrm{mg}$

", "

$\\mathrm{lb}$

", "

$\\mathrm{tonne}$

", "

$\\mathrm{gm}$

", "

$\\mathrm{K}$

", "

$\\mathrm{kg}$

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Choose the base SI unit for mass:

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$\\mathrm{c}$

", "

$\\mathrm{i}$

", "

$\\mathrm{V}$

", "

$\\mathrm{am}$

", "

$\\mathrm{r}$

", "

$\\mathrm{A}$

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Choose the base SI unit for electrical current:

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$\\mathrm{Ce}$

", "

$\\mathrm{F}$

", "

$\\mathrm{C}$

", "

$\\mathrm{T}$

", "

$\\mathrm{k}$

", "

$\\mathrm{K}$

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Choose the base SI unit for temperature:

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Identify the appropraite SI units for different measurements.

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Here is a table of base SI units:

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
QuantityUnitUnit symbol
lengthmetrem
masskilogramkg
timeseconds
electric currentampereA
temperaturekelvinK
amount of substancemolemol
luminous intensitycandelacd
\n

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Given the units displayed, choose the correct exponential value in terms of the given base unit (which is not necessarily the SI unit).

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$M\\var{x1}$

", "

$n\\var{x2}$

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$c\\var{x3}$

", "

$m\\var{x4}$

", "

$p\\var{x5}$

", "

$\\mu\\var{x6}$

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$\\times 10^{-15}$

", "

$\\times 10^{-12}$

", "

$\\times 10^{-9}$

", "

$\\times 10^{-6}$

", "

$\\times 10^{-3}$

", "

$\\times 10^{-2}$

", "

$\\times 10^{-1}$

", "

$\\times 10^{1}$

", "

$\\times 10^{2}$

", "

$\\times 10^{3}$

", "

$\\times 10^{6}$

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Match the SI prefix with the order of magnitude (e.g.,  'kg' means 10^3 of the base unit 'g')

"}, "preamble": {"css": "", "js": ""}, "functions": {}, "advice": "

Here is a table of metric prefixes:

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
petaP$10^{15}$
teraT$10^{12}$
gigaG$10^{9}$
megaM$10^{6}$
kilok$10^{3}$
hectoh$10^{2}$
decada$10^{1}$
decid$10^{-1}$
centic$10^{-2}$
millim$10^{-3}$
micro$\\mu$$10^{-6}$
nanon$10^{-9}$
picop$10^{-12}$
femtof$10^{-15}$
\n

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Some rules of the SI system:

\n\n\n\n\n\n\n\n
\n

Convert the following to the unit specified and express the final answers in exponential notation i.e. in the form $a \\times 10^b$ for suitable $a$ and $b$.

\n

If the answer is $1.1 \\times 10^3$ you input 1.1 in the first box and 10^3 in the second box.

\n

If the answer is $1.1 \\times 10^{-3}$ then you input 10^(-3) in the second box, including the brackets.

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\n

$\\var{y2}$ $\\mathrm{cm}$ = [[0]] $\\phantom{{}}\\times\\phantom{{}}$ [[1]] $\\mathrm{mm}$,

\n

$\\var{y2}$ $\\mathrm{cm}$ = [[2]] $\\phantom{{}}\\times\\phantom{{}}$ [[3]] $\\mathrm{m}$.

\n

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\n

You can click on steps to reveal more information on scientific notation and exponential numbers.

\n

$\\var{y1}$ $\\mu\\mathrm{mol}$ = [[0]] $\\phantom{{}}\\times\\phantom{{}}$ [[1]] $\\mathrm{nmol}$,

\n

$\\var{y1}$ $\\mu\\mathrm{mol}$ = [[2]] $\\phantom{{}}\\times\\phantom{{}}$ [[3]] $\\mathrm{mol}$.

\n

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Scientific Notation and Exponential Numbers.

\n \n \n \n

Exponential notation is an alternative method of expressing numbers.

\n \n \n \n

Exponential numbers take the form $a^n$, where $a$ is multiplied by itself $n$ times. A simple example is $8 = 2^3 = 2 \\times 2 \\times 2$.

\n \n \n \n

In exponential notation, $a$ is termed the base while $n$ is termed the power or exponent or index.

\n \n \n \n

Scientific notation is a specific example of exponential numbers, $10$ is almost always used as the base number. Thus $10^3$ means $10 \\times 10 \\times 10 = 1000$, while $10^{-3}$ is the notation for the reciprocal of $10^3$ namely $\\frac{1}{1000}$.

\n \n \n \n

The other name for this mathematical format is standard form (you may have come across this in GCSE mathematics).

\n \n \n \n

Expressing numbers which are not whole powers of $10$ in scientific notation often requires a further multiplier, termed the coefficient ($C$), giving the expression in the form $C \\times 10^n$.

\n \n \n \n

For example, $5 \\times 10^3$ is the scientific notation for the number $5000$, while $3.25 \\times 10^2$ is the scientific notation for the number $325$.

\n \n \n \n

Similarly, $3.25 \\times 10^{-2}$ would represent $3.25 \\times \\frac{1}{100} = 0.0325$.

\n \n \n \n

It is scientific convention to express all such numbers as coefficients between $1$ and $10$ followed by an appropriate power of $10$ (for example, as $3.25 \\times 10^{-2}$, not as $0.325 \\times 10^{-1}$, nor as $32.5 \\times 10^{-3}$).

\n \n \n \n

You should use such scientific notation whenever you express very large or very small numbers – it is a recognized form of “shorthand”, and it avoids spurious accuracy (e.g. writing $9$ $000$ $000$ suggests that the number is exactly $9$ million, in contrast to $9.0 \\times 10^6$ which suggests no such accuracy beyond the first decimal place of the coefficient).

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$\\var{y3} \\times 10^{\\var{y4}}$ $\\mathrm{mg}$ = [[0]] $\\phantom{{}}\\times\\phantom{{}}$ [[1]] $\\mathrm{ng}$,

\n

$\\var{y3} \\times 10^{\\var{y4}}$ $\\mathrm{mg}$ = [[2]] $\\phantom{{}}\\times\\phantom{{}}$ [[3]] $\\mathrm{g}$.

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Convert measurements between different scales of SI unit.

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a)

\n

To convert from $\\mathrm{cm}$ to $\\mathrm{mm}$ and then convert to $\\mathrm{m}$ units.

\n

a) First convert $\\var{y2}$ $\\mathrm{cm}$ to exponential notation:
$\\var{y2}$ $\\mathrm{cm}$ = $\\var{y2}/100 \\times 10^2$ $\\mathrm{cm}$.

\n

b) Now write $1$ $\\mathrm{cm}$ in terms of $\\mathrm{mm}$ units:
$1~\\mathrm{cm} = 10^1\\mathrm{mm} = 10~\\mathrm{mm}$.

\n

c) Putting these together gives:
$\\var{y2}$ $\\mathrm{cm}$ = $\\var{y2}/100 \\times 10^3$ $\\mathrm{mm}$.

\n

To convert from $\\mathrm{mm}$ to $\\mathrm{m}$ units we have to multiply by $10^{-3}$ and so
we get $\\var{y2/100} \\times 10^0$ $\\mathrm{m}$.

\n

b)

\n

To convert from $\\mu\\mathrm{mol}$ to $\\mathrm{nmol}$ and then convert to $\\mathrm{mol}$ units.

\n

a) First convert $\\var{y1}$ $\\mu\\mathrm{mol}$ to exponential notation:
$\\var{y1}$ $\\mu\\mathrm{mol}$ = $\\var{y1}/1000 \\times 10^3$ $\\mu\\mathrm{mol}$.

\n

b) Now write $1$ $\\mu\\mathrm{mol}$ in terms of $\\mathrm{nmol}$ units:
$1$ $\\mu\\mathrm{mol}$ = $10^3$ $\\mathrm{nmol}$.

\n

c) Putting these together gives:
$\\var{y1}$ $\\mu\\mathrm{mol}$ = $\\var{y1}/1000 \\times 10^6$ $\\mathrm{nmol}$.

\n

To convert from $\\mathrm{nmol}$ to $\\mathrm{mol}$ units we have to multiply by $10^{-9}$ and so we get $\\var{y1/1000} \\times 10^{-3}$ $\\mathrm{mol}$.

\n

c)

\n

$\\var{y3} \\times 10^{\\var{y4}}$ $\\mathrm{mg}$ is already in exponential notation and to convert to $\\mathrm{ng}$ units we use the fact that $1~\\mathrm{mg} = 10^6~\\mathrm{ng}$.

\n

So, $\\var{y3} \\times 10^{\\var{y4}}$ $\\mathrm{mg}$ = $\\var{y3} \\times 10^{\\var{y4 +6}}$ $\\mathrm{ng}$.

\n

To convert from $\\mathrm{ng}$ to grammes ($\\mathrm{g}$) we have to multiply by $10^{-9}$ to get $\\var{y3} \\times 10^{\\var{y4 - 3}}$ $\\mathrm{g}$.

"}, {"name": "Amount of substance needed to prepare a solution", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Bill Foster", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/6/"}, {"name": "Christian Lawson-Perfect", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/7/"}, {"name": "Chris Graham", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/369/"}], "statement": "

How would you prepare the following solutions?

\n

Note that you can click on Show steps for each of the following questions. There you can follow a sequence of steps to get the answer.

\n

However, you should be able (perhaps after practice) to answer these questions without using Show steps.

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$\\var{z1}~\\mathrm{ml}$ of a sucrose solution at a concentration $\\var{z2}~\\mathrm{g}/\\mathrm{l}$.

\n

Grams of sucrose needed = [[0]] $\\mathrm{g}$ in $\\var{z1}~\\mathrm{ml}$ (to $3$ decimal places).

", "steps": [{"extendBaseMarkingAlgorithm": true, "scripts": {}, "type": "information", "variableReplacementStrategy": "originalfirst", "customMarkingAlgorithm": "", "showFeedbackIcon": true, "prompt": "

In this question you are given the required volume of the solution, $\\var{z1}~\\mathrm{ml}$, and the required concentration, $\\var{z2}~\\mathrm{g}/\\mathrm{l}$. You need to calculate how many grams are required per $\\var{z1}~\\mathrm{ml}$ to achieve this concentration. 

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The first step is to express the required concentration in $\\mathrm{ml}$ by dividing by $1000$.

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Now multiply this by $\\var{z1}$ to obtain the amount of sucrose needed in the given volume of solution.

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You have not given your answer to the correct precision.

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$\\var{z3}~\\mathrm{ml}$ of a $\\var{z4}~\\mathrm{molar}$ (can be expressed as $\\mathrm{mol}/\\mathrm{l}$ or $\\mathrm{mol~l}^{-1}$, but usually expressed as $\\mathrm{M}$) solution of KCl (relative molecular mass $74.5$) in water.

\n

Grams of KCl needed = [[0]] $\\mathrm{g}$ in $\\var{z3}~\\mathrm{ml}$ (answer to $3$ decimal places).

", "steps": [{"extendBaseMarkingAlgorithm": true, "scripts": {}, "type": "information", "variableReplacementStrategy": "originalfirst", "customMarkingAlgorithm": "", "showFeedbackIcon": true, "prompt": "

The first step is to express $1~\\mathrm{mol}$ of KCl in grams.

\n

$1~\\mathrm{mol}$ solution = $74.5~\\mathrm{g}$ KCl in $1~\\mathrm{L}$

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Next, express $\\var{z4}~\\mathrm{mol}$ as grams of KCl by multiplying $74.5~\\mathrm{g}$ by $\\var{z4}$.

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Since $1~\\mathrm{L} = 1000~\\mathrm{ml}$ we can find the number of grams in $1\\mathrm{ml}$ by dividing by $1000$.

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Finally, to get the number of grams needed to give $\\var{z3}~\\mathrm{ml}$ of $\\var{z4}~\\mathrm{M}$ KCl solution you multiply this result by $\\var{z3}$. Give your answer here to 3 decimal places.

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$\\var{z5}~\\mathrm{L}$ of $\\var{z6}~\\mathrm{millimolar}$ ($\\mathrm{mmol}/\\mathrm{l}$, or $\\mathrm{mmol}^{-1}$ or $\\mathrm{mM}$) glucose (relative molecular mass $180.2$) in water.

\n

Grams of glucose needed = [[0]] $\\mathrm{g}$ in $\\var{z5}~\\mathrm{L}$ (answer to $3$ decimal places).

", "steps": [{"extendBaseMarkingAlgorithm": true, "scripts": {}, "type": "information", "variableReplacementStrategy": "originalfirst", "customMarkingAlgorithm": "", "showFeedbackIcon": true, "prompt": "

The first step is to express $1~\\mathrm{mol}$ of glucose in grams in $1~\\mathrm{L}$.

\n

$1~\\mathrm{mol}$ solution = $180.2~\\mathrm{g}$ glucose in $1~\\mathrm{L}$

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Multiply this by $\\var{z5}$ to obtain the number of grams of glucose in $\\var{z5}~\\mathrm{l}$ of $1~\\mathrm{mol}$ solution.

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Now you convert to $\\mathrm{mmol}$ by dividing by $1000$.

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So for $\\var{z6}~\\mathrm{mmol}$ we multiply by $\\var{z6}$ to get the required number of grams of glucose in $\\var{z5}~\\mathrm{L}$. Give your answer here to 3 decimal places.

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Preparing solutions to given concentrations/dilutions.

"}, "functions": {}, "advice": "

These question test your ability to convert between different SI units.

\n

a)

\n

Here, you are given the required volume of the solution, $\\var{z1}~\\mathrm{ml}$, and the required concentration, $3~\\mathrm{g}/\\mathrm{l}$. You need to calculate how many grams are required per $\\var{z1}~\\mathrm{ml}$. Therefore,

\n

$\\var{z2}~\\mathrm{g}/\\mathrm{l} = \\frac{\\var{z2}~\\mathrm{g}}{1000~\\mathrm{ml}} = \\frac{\\var{z2}~\\mathrm{g}}{\\var{1000/z1} \\times \\var{z1}~\\mathrm{ml}} = \\var{z1*z2/1000}~\\mathrm{g}$ in $\\var{z1}~\\mathrm{ml}$.

\n

b)

\n

This can be calculated in a similar way.

\n

$1~\\mathrm{mol}$ solution = $74.5$ KCl in $1~\\mathrm{L}$.

\n

$\\var{z4}~\\mathrm{mol}$ solution = $\\var{z4} \\times 74.5~\\mathrm{g}$ KCl in $1~\\mathrm{L}$.

\n

$\\var{z4}~\\mathrm{mol}$ solution = $\\var{z4*74.5}~\\mathrm{g}$ KCl in $1000~\\mathrm{ml}$.

\n

$\\var{z4}~\\mathrm{mol}$ solution = $\\frac{\\var{z4*74.5}}{1000}~\\mathrm{g}$ KCl in $1~\\mathrm{ml}$.

\n

$\\var{z4}~\\mathrm{mol}$ solution = $\\var{z3} \\times \\var{z4*74.5/1000}~\\mathrm{g}$ KCl in $\\var{z3}~\\mathrm{ml}$.

\n

$\\var{z4}~\\mathrm{mol}$ solution = $\\var{z8}~\\mathrm{g}$ KCl in $\\var{z3}~\\mathrm{ml}$.

\n

c)

\n

Again, this is similar.

\n

$1~\\mathrm{mol}$ solution = $180.2~\\mathrm{g}$ glucose in $1~\\mathrm{L}$.

\n

$1~\\mathrm{mol}$ solution = $\\var{z5} \\times 180.2~\\mathrm{g}$ glucose in $\\var{z5}~\\mathrm{L}$.

\n

$1000~\\mathrm{mmol}$ solution = $\\var{z5} \\times 180.2~\\mathrm{g}$ glucose in $\\var{z5}~\\mathrm{L}$.

\n

$1~\\mathrm{mmol}$ solution = $\\frac{\\var{z5} \\times 180.2}{1000}~\\mathrm{g}$ glucose in $\\var{z5}~\\mathrm{L}$.

\n

$\\var{z6}~\\mathrm{mmol}$ solution = $\\frac{\\var{z6} \\times \\var{z5} \\times 180.2}{1000}~\\mathrm{g}$ glucose in $\\var{z5}~\\mathrm{L}$.

\n

$\\var{z6}~\\mathrm{mmol}$ solution = $\\var{z9}~\\mathrm{g}$ glucose in $\\var{z5}~\\mathrm{L}$.

"}, {"name": "Calculate dilution and concentration factors", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Bill Foster", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/6/"}, {"name": "Chris Graham", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/369/"}], "tags": [], "variablesTest": {"condition": "", "maxRuns": 100}, "variables": {"w10": {"description": "", "name": "w10", "definition": "precround((0.5*(w8+w9)),1)", "templateType": "anything", "group": "Ungrouped variables"}, "w7": {"description": "", "name": "w7", "definition": "random('glucose', 'surcose', 'NaCl')", "templateType": "anything", "group": "Ungrouped variables"}, "w4": {"description": "", "name": "w4", "definition": "random(2..10#1)", "templateType": "anything", "group": "Ungrouped variables"}, "w5": {"description": "", "name": "w5", "definition": "(w1*w4)", "templateType": "anything", "group": "Ungrouped variables"}, "w14": {"description": "", "name": "w14", "definition": "precround(w10*w13/(w10+w11), 4)", "templateType": "anything", "group": "Ungrouped variables"}, "w3": {"description": "", "name": "w3", "definition": "(w1*w2)", "templateType": "anything", "group": "Ungrouped variables"}, "w2": {"description": "", "name": "w2", "definition": "random(3..8#1)", "templateType": "anything", "group": "Ungrouped variables"}, "w15": {"description": "", "name": "w15", "definition": "precround(w10*w13/w12, 4)", "templateType": "anything", "group": "Ungrouped variables"}, "w6": {"description": "", "name": "w6", "definition": "random(0.15..0.95#0.05)", "templateType": "anything", "group": "Ungrouped variables"}, "w8": {"description": "", "name": "w8", "definition": "random(2..9#1)", "templateType": "anything", "group": "Ungrouped variables"}, "w12": {"description": "", "name": "w12", "definition": "(w10 + w11 + random(5..10#1))", "templateType": "anything", "group": "Ungrouped variables"}, "w13": {"description": "", "name": "w13", "definition": "(w6*w8/(w9+w8))", "templateType": "anything", "group": "Ungrouped variables"}, "w1": {"description": "", "name": "w1", "definition": "random(5..20#1)", "templateType": "anything", "group": "Ungrouped variables"}, "w11": {"description": "", "name": "w11", "definition": "random(5..10#1)", "templateType": "anything", "group": "Ungrouped variables"}, "w9": {"description": "", "name": "w9", "definition": "random(3..12#1)", "templateType": "anything", "group": "Ungrouped variables"}}, "statement": "

There are many occasions when you have to work out what happens when solutions are diluted (or concentrated). The following relationship is useful, if you are diluting (or concentrating), it tells you that the mass remains constant:

\n

\\[\\begin{eqnarray} C_1V_1 = C_2V_2 \\end{eqnarray}\\]

\n

where you are diluting or concentrating volume $V_1$ with initial concentration $C_1$ to get volume $V_2$ with new concentration $C_2$.

\n

The important aspect to consider is either (a) the ratio of initial and final volumes used (this is often termed the 'dilution factor’) or (b) the ratio of initial and final concentrations required (depending upon the calculation).

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(i)

\n \n \n \n

If $\\var{w1}~\\mathrm{ml}$ of an aqueous sample is added to $\\var{w3}~\\mathrm{ml}$ of diluent (diluting liquid), what is the diluting factor?

\n \n \n \n

Dilution factor = $1$ in [[0]].

\n \n \n \n

(ii)

\n \n \n \n

If instead the $\\var{w1}~\\mathrm{ml}$ of an aqueous sample is made up to $\\var{w5}~\\mathrm{ml}$ by adding diluent, what is the diluting factor?

\n \n \n \n

Dilution factor = $1$ in [[1]].

\n \n ", "steps": [{"extendBaseMarkingAlgorithm": true, "scripts": {}, "type": "information", "variableReplacementStrategy": "originalfirst", "customMarkingAlgorithm": "", "showFeedbackIcon": true, "prompt": "\n \n \n

For part (i) you take the ratio of the final volume to the initial volume to get the dilution factor. The initial volume is $\\var{w1}~\\mathrm{ml}$. The final volume is $\\var{w1} + \\var{w3} = \\var{w1 + w3}~\\mathrm{ml}$ and so you can calculate the dilution factor.

\n \n \n \n

In part (ii) the final volume is now $\\var{w5}~\\mathrm{ml}$ and you use the same formula for the dilution factor.

\n \n ", "useCustomName": false, "variableReplacements": [], "marks": 0, "customName": "", "adaptiveMarkingPenalty": 0, "showCorrectAnswer": true, "unitTests": []}], "useCustomName": false, "variableReplacements": [], "marks": 0, "customName": "", "adaptiveMarkingPenalty": 0, "showCorrectAnswer": true, "unitTests": []}, {"extendBaseMarkingAlgorithm": true, "scripts": {}, "stepsPenalty": 0, "type": "gapfill", "variableReplacementStrategy": "originalfirst", "sortAnswers": false, "customMarkingAlgorithm": "", "gaps": [{"extendBaseMarkingAlgorithm": true, "precisionMessage": "You have not given your answer to the correct precision.", "precisionPartialCredit": 0, "scripts": {}, "type": "numberentry", "variableReplacementStrategy": "originalfirst", "adaptiveMarkingPenalty": 0, "showFeedbackIcon": true, "allowFractions": false, "mustBeReduced": false, "minValue": "{w14 - 0.0001}", "variableReplacements": [], "marks": 1, "strictPrecision": false, "showCorrectAnswer": true, "unitTests": [], "correctAnswerFraction": false, "notationStyles": ["plain", "en", "si-en"], "mustBeReducedPC": 0, "precision": "4", "customMarkingAlgorithm": "", "maxValue": "{w14 + 0.0001}", "precisionType": "dp", "correctAnswerStyle": "plain", "useCustomName": false, "showPrecisionHint": false, "customName": ""}, {"extendBaseMarkingAlgorithm": true, "precisionMessage": "You have not given your answer to the correct precision.", "precisionPartialCredit": 0, "scripts": {}, "type": "numberentry", "variableReplacementStrategy": "originalfirst", "adaptiveMarkingPenalty": 0, "showFeedbackIcon": true, "allowFractions": false, "mustBeReduced": false, "minValue": "{w15 - 0.0001}", "variableReplacements": [], "marks": 1, "strictPrecision": false, "showCorrectAnswer": true, "unitTests": [], "correctAnswerFraction": false, "notationStyles": ["plain", "en", "si-en"], "mustBeReducedPC": 0, "precision": "4", "customMarkingAlgorithm": "", "maxValue": "{w15 + 0.0001}", "precisionType": "dp", "correctAnswerStyle": "plain", "useCustomName": false, "showPrecisionHint": false, "customName": ""}], "showFeedbackIcon": true, "prompt": "

You have a solution containing $\\var{w6}~\\mathrm{mol}/\\mathrm{l}$ {w7} and you add $\\var{w8}~\\mathrm{ml}$ of this solution to $\\var{w9}~\\mathrm{ml}$ of sterile water in a test tube.

\n

(i)

\n

After mixing these liquids together, you take $\\var{w10}~\\mathrm{ml}$ and add it to a conical flask containing $\\var{w11}~\\mathrm{ml}$ of sterile water. What would be the final concentration of {w7} (in $\\mathrm{mol}/\\mathrm{l}$) in the conical flask at the end of the procedure?

\n

Final concentration = [[0]] $\\mathrm{M}$     give your answer to 4 decimal places

\n

(ii)

\n

Alternatively, after mixing the liquids you take $\\var{w10}~\\mathrm{ml}$ as before and make it up to $\\var{w12}~\\mathrm{ml}$ in a conical flask by adding sterile water. What now would be the final concentration of {w7} (in $\\mathrm{mol}/\\mathrm{l}$) in the conical flask at the end of this procedure?

\n

Final concentration = [[1]] $\\mathrm{M}$       give your answer to 4 decimal places

", "steps": [{"extendBaseMarkingAlgorithm": true, "scripts": {}, "type": "information", "variableReplacementStrategy": "originalfirst", "customMarkingAlgorithm": "", "showFeedbackIcon": true, "prompt": "\n \n \n

(i) You do this in two steps. Note that initially you have $C_1 = \\var{w6}~\\mathrm{mol}/\\mathrm{l}$ and $V_1 = \\var{w8}~\\mathrm{ml}$.

\n \n \n \n

Step 1. Find the concentration $C_2$ after adding the $\\var{w8}~\\mathrm{ml}$ of the solution to the $\\var{w9}~\\mathrm{ml}$ of sterile water so that $V_2 = \\var{w8} + \\var{w9} = \\var{w8 + w9}~\\mathrm{ml}$. You use $C_1V_1 = C_2V_2$ to find $C_2$.

\n \n \n \n

Step 2. Now you take $V_3 = \\var{w10}~\\mathrm{ml}$ at the concentration $C_2$ you have just found and add it to $\\var{w11}~\\mathrm{ml}$ of sterile water to get a volume $V_4 = \\var{w10} + \\var{w11} = \\var{w10 + w11}~\\mathrm{ml}$. If $C_4$ is the concentration of this volume you use $C_2V_3 = C_4V_4$ and so find $C_4~\\mathrm{mol}/\\mathrm{l}$ which is the final concentration required.

\n \n \n \n

(ii) As in part (i) above, you split the problem into two steps. You get the same concentration at the end of Step 1. The only difference is that now $V_4 = \\var{w12}$ but you use the same formula $C_2V_3 = C_4V_4$ to find $C_4$.

\n \n ", "useCustomName": false, "variableReplacements": [], "marks": 0, "customName": "", "adaptiveMarkingPenalty": 0, "showCorrectAnswer": true, "unitTests": []}], "useCustomName": false, "variableReplacements": [], "marks": 0, "customName": "", "adaptiveMarkingPenalty": 0, "showCorrectAnswer": true, "unitTests": []}], "ungrouped_variables": ["w15", "w14", "w13", "w12", "w11", "w10", "w7", "w6", "w5", "w4", "w3", "w2", "w1", "w9", "w8"], "rulesets": {}, "preamble": {"css": "", "js": ""}, "metadata": {"licence": "Creative Commons Attribution 4.0 International", "description": "

Questions on dilutions and concentrations in making up aqueous solutions.

"}, "functions": {}, "advice": "

a)

\n

For part (i) you take the ratio of the final volume to the initial volume to get the dilution factor.

\n

The initial volume is $\\var{w1}~\\mathrm{ml}$.

\n

The final volume is $\\var{w1} + \\var{w3} = \\var{w1 + w3}~\\mathrm{ml}$ and so the dilution factor is $\\frac{\\var{w1 + w3}}{\\var{w1}} = \\var{(w1+w3)/w1}$.

\n

In part (ii) the final volume is now $\\var{w5}~\\mathrm{ml}$ and you use the same formula for the dilution factor giving $\\frac{\\var{w5}}{\\var{w1}} = \\var{w5/w1}$.

\n

b)

\n

(i) You do this in two steps. Note that initially you have $C_1 = \\var{w6}~\\mathrm{M}$ and $V_1 = \\var{w8}~\\mathrm{ml}$.

\n

Step 1. Find the concentration $C_2$ after adding the $\\var{w8}~\\mathrm{ml}$ of the solution to the $\\var{w9}~\\mathrm{ml}$ of sterile water so that $V_2 = \\var{w8} + \\var{w9} = \\var{w8 + w9}~\\mathrm{ml}$.

\n

This is given by $C_2 = \\frac{\\var{w8} \\times \\var{w6}}{\\var{w8} + \\var{w9}} = \\frac{\\var{w8*w6}}{\\var{w8+w9}} = \\var{precround(w13,4)}~\\mathrm{M}$ to $4$ decimal places.

\n

Step 2. Now you take $V_3 = \\var{w10}~\\mathrm{ml}$ at the concentration $C_2 = \\var{precround((w8*w6)/(w8+w9),4)}~\\mathrm{M}$ and add it to $\\var{w11}~\\mathrm{ml}$ of sterile water to get a volume $V_4 = \\var{w10} + \\var{w11} = \\var{w10 + w11}~\\mathrm{ml}$. If $C_4$ is the concentration of this volume you use $C_2V_3 = C_4V_4$ and so $C_4 = \\frac{C_2V_3}{V_4} = \\frac{\\var{precround((w8*w6)/(w8+w9),4)} \\times \\var{w10}}{\\var{w10 + w11}} = \\var{precround((w8*w6*w10)/((w8+w9)*(w10+w11)),4)}~\\mathrm{M}$ which is the final concentration required.

\n

(ii) As in part a) above, you split the problem into two steps. You get the same concentration at the end of Step 1. The only difference is that now $V_4 = \\var{w12}$ and so in this case $C_4 = \\frac{C_2V_3}{V_4} = \\frac{\\var{precround((w8*w6)/(w8+w9),4)} \\times \\var{w10}}{\\var{w12}} = \\var{precround(w15,4)}~\\mathrm{M}$.

"}, {"name": "Concentrations in molarity", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Bill Foster", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/6/"}, {"name": "Christian Lawson-Perfect", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/7/"}, {"name": "Chris Graham", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/369/"}], "statement": "

The following questions ask you to express concentration in different ways.

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Given concentration expressed in molarity (M).

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Density of ethanol at 25C

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Relative molecular mass of ethanol

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Concentration of ethanol (% w/v)

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What is $\\var{a1}~\\mathrm{gl}^{-1}$ sucrose expressed in terms of molarity (relative molecular mass of sucrose is $\\var{rmm1}$)?

\n

[[0]] $\\mathrm{mmol}$       give you anwer to 2 decimal places

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You must express $\\mathrm{gl}^{-1}$ in terms of molarity. You
are given that the relative molecular mass of sucrose is $\\var{rmm1}$, and this
tells us that $\\var{rmm1}~\\mathrm{gl}^{-1}$ sucrose $= 1~\\mathrm{mol}$.

\n

In order to calculate $\\var{a1}~\\mathrm{gl}^{-1}$ sucrose in terms of molarity we first find it for $1~\\mathrm{gl}^{-1}$ and then for $\\var{a1}~\\mathrm{gl}^{-1}$ by multiplying by $\\var{a1}$ and following that convert to $\\mathrm{mmol}$.

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First, write $1~\\mathrm{gl}^{-1}$ sucrose in $\\mathrm{mol}$. Enter your answer to 4 decimal places.

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You have not given your answer to the correct precision.

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Next, mutliply by $\\var{a1}$ to obtain $\\var{a1}~\\mathrm{gl}^{-1}$ sucrose in $\\mathrm{mol}$. Enter your answer to 3 decimal places.

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You have not given your answer to the correct precision.

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Finally, convert to $\\mathrm{mmol}$ by multiplying by $1000$. Enter your answer to 2 decimal places.

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What is $\\var{a2}~\\mathrm{M}$ NaCl expressed in $\\mathrm{gl}^{-1}$ (relative molecular mass of NaCl is $\\var{rmm2}$)?

\n

[[0]] $\\mathrm{gl}^{-1}$     give you anwer to 3 decimal places

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You must express $\\mathrm{M}$ in terms of $\\mathrm{gl}^{-1}$.

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Using the information given on the relative mass of NaCl you can write $1~\\mathrm{M}$ NaCl in $\\mathrm{gl}^{-1}$.

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Now multiply by $\\var{a2}$ to get the result. Give your answer to 3 decimal places.

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You have not given your answer to the correct precision.

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What is $\\var{a3}~\\%\\mathrm{v}/\\mathrm{v}$ ethanol expressed in terms of molarity? Express your answer in $\\mathrm{M}$, which is the same as $\\mathrm{mol}/\\mathrm{l}$ or 'moles per litre'. (The relative molecular mass of ethanol is $\\var{rmm3}$ and density of ethanol at $25^{\\circ}C$ is $\\var{d1}~\\mathrm{gml}^{-1}$)?

\n

[[0]] $\\mathrm{M}$        give you anwer to 3 decimal places

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You must express $\\%\\mathrm{v}/\\mathrm{v}$ in terms of molarity. The strategy is to find the mass of ethanol in one litre given the concentration and then find the mass of this volume in $\\mathrm{g}$. Concentration can then be converted into units of molarity.

\n

Fill in the following:

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$\\var{a3}\\%$ ethanol contains how much ethanol per litre, in $\\mathrm{ml}$?

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What is the mass of this ethanol in $\\mathrm{g}$? (Here use the density $\\var{d1}~\\mathrm{gml}^{-1}$)

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Now $\\var{rmm3}~\\mathrm{g}$ of ethanol in $1~\\mathrm{L} = 1~\\mathrm{M}$.

\n

Hence:

\n

$1~\\mathrm{g}$ of ethanol in $1~\\mathrm{L} = \\frac{1}{\\var{rmm3}}~\\mathrm{M}$

\n

Multiply by the mass of the ethanol in $1$ litre, calculated above, to find the concentration in terms of $\\mathrm{M}$.

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Expressing concentrations of solutions in different ways.

"}, "advice": "

a)

\n

Here, you must express $\\mathrm{gl}^{-1}$ in terms of molarity. Therefore:

\n

$\\var{rmm1}~\\mathrm{gl}^{-1}$ sucrose = $1~\\mathrm{mol}$.

\n

$1~\\mathrm{gl}^{-1}$ sucrose = $\\frac{1}{\\var{rmm1}}~\\mathrm{mol}$.

\n

$\\var{a1}~\\mathrm{gl}^{-1}$ sucrose = $\\frac{\\var{a1}}{\\var{rmm1}}~\\mathrm{mol}$.

\n

$\\var{a1}~\\mathrm{gl}^{-1}$ sucrose = $\\frac{\\var{1000*a1}}{\\var{rmm1}}~\\mathrm{mmol}$.

\n

$\\var{a1}~\\mathrm{gl}^{-1}$ sucrose = $\\var{precround(ans1,2)}~\\mathrm{mmol}$ to $2$ decimal places.

\n

b)

\n

Here, you must express $\\mathrm{M}$ in terms of $\\mathrm{gl}^{-1}$. Therefore:

\n

$1~\\mathrm{M}$ NaCl = $\\var{rmm2}~\\mathrm{gl}^{-1}$.

\n

$\\var{a2}~\\mathrm{M}$ NaCl = $\\var{a2} \\times \\var{rmm2}~\\mathrm{gl}^{-1}$.

\n

$\\var{a2}~\\mathrm{M}$ NaCl = $\\var{precround(ans2,3)}~\\mathrm{gl}^{-1}$ to $3$ decimal places.

\n

c)

\n

Here, you must express $\\%\\mathrm{w}/\\mathrm{v}$ in terms of molarity. Therefore:

\n

$\\var{a3}\\%$ ethanol contains $\\var{a3}~\\mathrm{ml}$ ethanol per $100~\\mathrm{ml}$ solution, or $100~\\mathrm{ml}$ ethanol in a litre.

\n

$\\var{10*a3}~\\mathrm{ml}$ ethanol weighs $\\var{10*a3} \\times \\var{d1}~\\mathrm{g} = \\var{w1}~\\mathrm{g}$.

\n

$\\var{rmm3}~\\mathrm{g}$ of ethanol in $1~\\mathrm{L} = 1~\\mathrm{mol}$.

\n

$1~\\mathrm{g}$ of ethanol in $1~\\mathrm{L} = \\frac{1}{\\var{rmm3}}~\\mathrm{mol}$.

\n

$\\var{w1}~\\mathrm{g}$ of ethanol in $1~\\mathrm{L} = \\frac{\\var{w1}}{\\var{rmm3}}~\\mathrm{mol} = \\var{precround(ans3,3)}~\\mathrm{mol}$ to $3$ decimal places.

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