# Question 816b8

Dec 5, 2014

After making this more of a sample problem, I believe the answer is $1.14 \frac{m o l}{{\mathrm{dm}}^{3}}$ (assuming 100 grams of water).

The solution's percent concentration by mass is defined as

c% = (mass_(solute))/(mass_(t otal solution)) * 100, where

$m a s {s}_{t o t a l s o l u t i o n} = m a s {s}_{s o l v e n t} + m a s {s}_{s o l u t e}$

Now, let us assume that we have $m a s {s}_{s o l v e n t} = 100 g$ of water, ${H}_{2} O$,

c% = (mass_(solute))/(mass_(solute) + 100) * 100# = $21.80$.

We would then have

$m a s {s}_{s o l u t e} = 0.218 \cdot \left(m a s {s}_{s o l u t e} + 100\right)$, which means

$m a s {s}_{s o l u t e} = 27.9 g r a m s$, the solute being of course ${K}_{2} C r {O}_{4}$.

The number of moles of ${K}_{2} C r {O}_{4}$ can be determined by

${n}_{{K}_{2} C r {O}_{4}} = \frac{27.9 g}{194.2 \frac{g}{m o l e}} = 0.14 m o l e s$, where $194.2 \frac{g}{m o l e}$ is the molar mass of the solute.

Knowing the density of the solute, we can determine its volume

${V}_{{K}_{2} C r {O}_{4}} = \frac{m}{\mathrm{de} n s i t y} = \frac{27.9 g}{1.20 \frac{g}{c {m}^{3}}} = 23.3 c {m}^{3}$

Knowing that $1 {\mathrm{dm}}^{3}$ = ${10}^{3} c {m}^{3}$, we get a volume of $V = 0.023 {\mathrm{dm}}^{3}$.

The density of water is $1 \frac{g}{c {m}^{3}}$, therefore the volume of water is equal to

${V}_{{H}_{2} O} = \frac{100 g}{1 \frac{g}{c {m}^{3}}} = 100 c {m}^{3} = 0.1 {\mathrm{dm}}^{3}$

Therefore, the molarity can be calculated by

molarity $= \frac{{n}_{{K}_{2} C r {O}_{4}}}{{V}_{t o t a l}} = \frac{0.14}{0.023 + 0.1} = 1.14 M$