Playing with Concentration 3

Chemistry Level 2

The molality of an alcohol $(\ce{C_2{H}_6O})$ solution is $3\text{ mol kg}^{-1}$ , and its density is $1.6\text{ g mL}^{-1}.$

At $4 ^\circ \text{C}$ , if I want to reduce half the mass percentage of the solution, how much water should I add in $\text{mL}$ ?

The answer is 1138.

1 solution

Christopher Boo
Apr 11, 2014

From Playing with Concentration 1 , we know the mass percentage is $12.127\%$ .

Reduce half the mass percentage of the solution will be $6.0635\%$ .

Let the mass of water added be $x$ ,

$\displaystyle \frac{138}{1138+x}\times 100\%=6.0635\%$

Then we will have $x=1138g$ and we know that the density of water at $4 ^\circ C$ is $1\text{ gmL}^{-1}$ .

Hence, the volume of water added is $1138\text{ mL}$ .

The answer depends on how much of the solution we have. If we have 1138 grams, then your answer is correct. However if we have 276 grams of ethanol in 2 kg of water, then the answer would be 2276 g. Could you please clarify your problem a little bit more?

Aditya Virani Staff - 5 years, 6 months ago

Playing with Concentration 3

The answer is 1138.

1 solution

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