# Question 8f15a

Feb 24, 2015

I think you've mistyped the hydrate, since the name suggests you'd have $K A l {\left(S {O}_{4}\right)}_{2} \cdot 12 {H}_{2} O$, not $K C l {\left(S {O}_{4}\right)}_{2} \cdot 12 {H}_{2} O$.

You'd need $\text{9.504 g}$ of potassium aluminium sulfate dodecahydrate to prepare that particular solution.

When you have problems like this one, the approach you have to take is to work backwards from the final solution to determine the mass of solute, in your case $K A l {\left(S {O}_{4}\right)}_{2}$, or potassium alum, you have in $\text{35.00 g}$ of the $\text{15.00%}$ solution.

What a $\text{15.00%}$ concentration by mass solutions means is that you have 15.00 g of apotassium alum for every 100 g of solution. So,

$\text{%15.00" = m_("potassium alum")/m_("solution") * 100 = m_("potassium alum")/"35.00 g} \cdot 100$

m_("potassium alum") = (15.00 * 35.00)/100 = "5.250 g" $K A l {\left(S {O}_{4}\right)}_{2}$

Next you need to determine how much hydrate you need to get this much potassium alum. To do that, determine the mass percent potassium alum has in the hydrate.

$\text{266.6 g/mol"/("482.6 g/mol") * 100 = "55.24%}$, where

$\text{266.6 g/mol}$ - the molar mass of potassium alum;
$\text{482.6 g/mol}$ - the molar mass of the hydrate;

So, for every 100 g of hydrate you get 55.24 g of potassium alum. This means that

$\text{5.250 g potassium alum" * ("100 g hydrate")/("55.24 g potassium alum") = "9.504 g hydrate}$

You'd get the needed 5.250 g of potassium alum from 9.504 g of hydrate. The mass of water you need to add to make the solution will be

${m}_{\text{water added") = m_("solution") - m_("hydrate}}$

m_("water added") = 35.00 - 9.504 = "25.50 g added water"

SIDE NOTE Keep in mind that the total mass of water in the solution is actually the sum of the water that comes with the hydrate and the mass of the added water

${m}_{\text{water total") = m_("water hydrate") + m_("water added}}$

m_("water total") = (9.504 - 5.250) + 25.50 = "29.75 g water"#