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How would you prepare the following solutions?

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.

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 $\\var{z2}~\\mathrm{g}/\\mathrm{l}$.

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

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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.

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

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

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

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.

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

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

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

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.

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

a)

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,

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

b)

This can be calculated in a similar way.

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

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

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

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

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

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

c)

Again, this is similar.

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

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

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

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

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

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

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