Question #20474

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Question #20474

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Answer 1 · 2014-12-10T20:14:30

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

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%.

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

We know that $m_{s o l u t i o n 1} = 70 g$ $m_(solution1) = 70g$ and the the concentration is equal to $c_{1} = 10 %$ $c_1 = 10%$ , which means we can determine the mass of $K C l$ $KCl$ used for the mixture

$m_{K C l} = \frac{c_{1} \cdot m_{s o l u t i o n}}{100} = \frac{10 \cdot 70}{100} = 7 g$ $m_(KCl) = (c_1 * m_(solution))/100 = (10 * 70)/100 = 7g$

Now, let's say that after adding a certain mass of water - $m_{a d d e d}$ $m_(added)$ - we would get a new concentration of $c_{2} = 2 %$ $c_2 = 2%$ . This means that

$c_{2} = \frac{m_{K C l}}{m_{s o l u t i o n 2}} \cdot 100$ $c_2 = m_(KCl)/m_(solution2) * 100$ , where

$m_{s o l u t i o n 2} = m_{s o l u t i u o n 1} + m_{a d d e d}$ $m_(solution2) = m_(solutiuon1) + m_(added)$

Therefore, $m_{s o l u t i o n 2} = \frac{m_{K C l} \cdot 100}{c_{2}} = 7 \cdot \frac{100}{2} = 350 g$ $m_(solution2) = (m_(KCl) * 100)/c_2 = 7 * 100/2 = 350g$

This means that $m_{a d d e d} = m_{s o l u t i o n 2} - m_{s o l u t i o n 1} = 350 - 70 = 280 g$ $m_(added) = m_(solution2) - m_(solution1) = 350 - 70 =280g$

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

$V_{w a t e r} = \frac{m_{a d d e d}}{ρ} = \frac{280 g}{1 \frac{g}{c m^{3}}} = 280 c m^{3}$ $V_(water) = m_(added)/rho = (280g)/(1 g/(cm^3)) = 280 cm^3$

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Question #20474

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