# Question 20474

Dec 10, 2014

The answer is $280 c {m}^{3}$.

A solution's percent concentration by mass is defined as the mass of the solute divided by the mass of the solution and multiplied by 100%.

c = m_(solute)/m_(solution) * 100%, where ${m}_{s o l u t i o n} = {m}_{s o l v e n t} + {m}_{s o l u t e}$.

We know that ${m}_{s o l u t i o n 1} = 70 g$ and the the concentration is equal to c_1 = 10%, which means we can determine the mass of $K C l$ used for the mixture

${m}_{K C l} = \frac{{c}_{1} \cdot {m}_{s o l u t i o n}}{100} = \frac{10 \cdot 70}{100} = 7 g$

Now, let's say that after adding a certain mass of water - ${m}_{a \mathrm{dd} e d}$ - we would get a new concentration of c_2 = 2%#. This means that

${c}_{2} = {m}_{K C l} / {m}_{s o l u t i o n 2} \cdot 100$, where

${m}_{s o l u t i o n 2} = {m}_{s o l u t i u o n 1} + {m}_{a \mathrm{dd} e d}$

Therefore, ${m}_{s o l u t i o n 2} = \frac{{m}_{K C l} \cdot 100}{c} _ 2 = 7 \cdot \frac{100}{2} = 350 g$

This means that ${m}_{a \mathrm{dd} e d} = {m}_{s o l u t i o n 2} - {m}_{s o l u t i o n 1} = 350 - 70 = 280 g$

Assuming we're at room temperature, we can determine the volume of water by using its known density of $\rho = 1 \frac{g}{c {m}^{3}}$

${V}_{w a t e r} = {m}_{a \mathrm{dd} e d} / \rho = \frac{280 g}{1 \frac{g}{c {m}^{3}}} = 280 c {m}^{3}$