# Question eed53

Dec 21, 2014

The answer is $950 c {m}^{3}$ of water must be added.

Let's start with the initial solution. We know that we have an initial volume of ${V}_{1} = 300.0 c {m}^{3}$ and a density of $\rho = 1.38 \frac{g}{c {m}^{3}}$. Automatically, we can find out what the mass of the solution is by using

$\rho = {m}_{s o l u t i o n} / {V}_{1} \to {m}_{s o l u t i o n} = \rho \cdot {V}_{1} = 1.38 \frac{g}{c {m}^{3}} \cdot 300.0 c {m}^{3} = 414 g$

We know that we're dealing with a 48% percent concentration by mass solution of ${H}_{2} S {O}_{4}$, which means that we have $48 g$ of ${H}_{2} S {O}_{4}$ for every $100 g$ of solution. This means that the mass of sufuric acid, which represents the solute is equal to

c = m_(solute)/m_(solution) * 100%

m_(solute) = (c * m_(solution))/(100%) = (48% * 414g)/(100%) = 198g#

The number of moles of sulfuric acid, knowing its molar mass of $98.0 \frac{g}{m o l}$, is equal to

${n}_{{H}_{2} S {O}_{4}} = \frac{m}{m o l a r m a s s} = \frac{198 g}{98.0 \frac{g}{m o l}} = 2.02$ moles

The acid's molarity, expressed in moles per ${\mathrm{dm}}^{3}$, is equal to

${C}_{1} = \frac{n}{V} = \frac{2.02 m o l e s}{300.0 \cdot {10}^{- 3} {\mathrm{dm}}^{3}} = 6.73 \frac{m o l e s}{{\mathrm{dm}}^{3}}$

We can now use a simple dilution calculation to determine the volume of the desired solution

${C}_{1} \cdot {V}_{1} = {C}_{2} \cdot {V}_{2} \to {V}_{2} = {C}_{1} / {C}_{2} \cdot {V}_{1}$

${V}_{2} = \frac{6.73}{1.61} \cdot 300.0 c {m}^{3} = 1250 c {m}^{3}$

This means that the volume of water that needs to be added is $1250 - 300 = 950 c {m}^{3}$.
You can check the answer by determining the second solution's density

${\rho}_{2} = {m}_{s o l u t i o n 2} / {V}_{2} = \frac{414 g + 950 g}{1250 c {m}^{3}} \approx 1.10 \frac{g}{c {m}^{3}}$ - what is given for the second solution.