Question #816b8

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Question #816b8

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Answer 1 · 2014-12-05T11:02:20

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

The solution's percent concentration by mass is defined as

$c % = \frac{m a s s_{s o l u t e}}{m a s s_{t o t a l s o l u t i o n}} \cdot 100$ $c% = (mass_(solute))/(mass_(t otal solution)) * 100$ , where

$m a s s_{t o t a l s o l u t i o n} = m a s s_{s o l v e n t} + m a s s_{s o l u t e}$ $mass_(t otal solution) = mass_(solvent) + mass_(solute)$

Now, let us assume that we have $m a s s_{s o l v e n t} = 100 g$ $mass_(solvent) = 100g$ of water, $H_{2} O$ $H_2O$ ,

$c % = \frac{m a s s_{s o l u t e}}{m a s s_{s o l u t e} + 100} \cdot 100$ $c% = (mass_(solute))/(mass_(solute) + 100) * 100$ = $21.80$ $21.80$ .

We would then have

$m a s s_{s o l u t e} = 0.218 \cdot (m a s s_{s o l u t e} + 100)$ $mass_(solute) = 0.218 * (mass_(solute) + 100)$ , which means

$m a s s_{s o l u t e} = 27.9 g r a m s$ $mass_(solute) = 27.9grams$ , the solute being of course $K_{2} C r O_{4}$ $K_2CrO_4$ .

The number of moles of $K_{2} C r O_{4}$ $K_2CrO_4$ can be determined by

$n_{K_{2} C r O_{4}} = \frac{27.9 g}{194.2 \frac{g}{m o l e}} = 0.14 m o l e s$ $n_(K_2CrO_4) = (27.9 g) / (194.2 g/(mo l e)) = 0.14 mol es$ , where $194.2 \frac{g}{m o l e}$ $194.2 g/(mo l e)$ is the molar mass of the solute.

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

$V_{K_{2} C r O_{4}} = \frac{m}{d e n s i t y} = \frac{27.9 g}{1.20 \frac{g}{c m^{3}}} = 23.3 c m^{3}$ $V_(K_2CrO_4) = m/(density) = (27.9 g)/(1.20 g/(cm^3)) = 23.3 cm^3$

Knowing that $1 d m^{3}$ $1 dm^3$ = $10^{3} c m^{3}$ $10^(3) cm^3$ , we get a volume of $V = 0.023 d m^{3}$ $V = 0.023 dm^3$ .

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

$V_{H_{2} O} = \frac{100 g}{1 \frac{g}{c m^{3}}} = 100 c m^{3} = 0.1 d m^{3}$ $V_(H_2O) = (100g)/(1 g/(cm^3)) = 100 cm^3 = 0.1 dm^3$

Therefore, the molarity can be calculated by

molarity $= \frac{n_{K_{2} C r O_{4}}}{V_{t o t a l}} = \frac{0.14}{0.023 + 0.1} = 1.14 M$ $= (n_(K_2CrO_4))/(V_(t otal)) = 0.14/(0.023+0.1) = 1.14 M$

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Question #816b8

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