Question #ade66

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Question #ade66

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Answer 1 · 2014-12-12T09:46:20

The answer is $2.15 g$ $2.15g$ .

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

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

Therefore, the mass of solute needed for this solution is

$m_{K_{2} C O_{3}} = \frac{c \cdot m_{s o l u t i o n}}{100} = \frac{5.00 % \cdot 43.0}{100 %} = 2.15 g$ $m_(K_2CO_3) = (c * m_(solution))/100 = (5.00% * 43.0)/(100%) = 2.15g$

It is important to try and estimate what that mass could be even before calculating it; this could have been done by takin the $\frac{1}{20}$ $1/20$ th part (the equivalent of 5%) of $43 g$ $43g$ , which is roughly $2 g$ $2g$ .

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Question #ade66

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