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

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

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.

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.