# How much 1.7M solution of (NH_4)_2SO_4 do you have to use so that it contains 0.230 g of the substance?

May 15, 2018

$\text{0.0011 L}$.

#### Explanation:

The mass of $1$ mole of ${\left(N {H}_{4}\right)}_{2} S {O}_{4}$ is:

$N \times 2 + H \times 4 \times 2 + S + O \times 4$
$= 14.01 \times 2 + 1.008 \times 4 \times 2 + 32.07 + 16.00 \times 4$
$= \text{132.14 g/mol}$

So, the number of moles of ${\left(N {H}_{4}\right)}_{2} S {O}_{4}$ in $\text{0.230 g}$ would be:

$\text{moles" = "mass"/"molar mass" = "0.230 g"/"132.14 g/mol" = "0.00187 mol}$

The question also tells us that the ${\left(N {H}_{4}\right)}_{2} S {O}_{4}$ solution is $\text{1.7 M}$. This means that there are 1.7" moles per liter.

$\text{1.7 moles"/"1 L}$

We can use this ratio to find the volume of solution which would have $\text{0.00187 mol}$ of ${\left(N {H}_{4}\right)}_{2} S {O}_{4}$:

$\text{1.7 moles"/"1 L" = "1.7 × (0.00187/1.7) moles"/"1.0 × (0.00187/1.7) L}$

$= \text{0.00187 moles"/"0.0011 L}$

We'd need $\text{0.0011 L}$ of the ${\left(N {H}_{4}\right)}_{2} S {O}_{4}$ solution to get $\text{0.230 g}$ of ${\left(N {H}_{4}\right)}_{2} S {O}_{4}$. :)