Playing with Concentration 4

Chemistry Level 2

The molality of an alcohol ( C X 2 H X 6 O ) (\ce{C_2{H}_6{O}}) solution ( A 1 ) (A_1) is 3 mol kg − 1 3\text{ mol kg}^{-1} , and its density is 1.6 g mL − 1 . 1.6\text{ g mL}^{-1}.

Now we have another alcohol solution ( A 2 ) (A_2) whose mass percentage is 98 % 98\% and density is 3.6 g mL − 1 . 3.6\text{ g mL}^{-1}. If we want to increase the mass percentage of alcohol in ( A 1 ) (A_1) by 3 times, what is the volume of the new alcohol solution A 2 A_2 we should add to it in mL ? \text{mL}?

Give your answer to 3 decimal places.


The answer is 124.428.

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Christopher Boo by Christopher Boo , 24, Singapore

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1 solution

Christopher Boo
Apr 11, 2014

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

Increase the mass percentage of the alcohol solution by three times will be 36.381 % 36.381\% .

Let the mass of the new alcohol solution added be x x .

138 + 0.98 x 1138 + x × 100 % = 36.381 % \displaystyle \frac{138+0.98x}{1138+x}\times 100\%=36.381\%

x = 447.939 g x=447.939g

Given that the density of ( A 2 ) (A_2) is 3.6 gmL − 1 3.6\text{ gmL}^{-1} , then the volume of ( A 2 ) (A_2) added is

447.939 3.6 = 124.428 mL \frac{447.939}{3.6}=124.428\text{ mL} .

1 Helpful 0 Interesting 0 Brilliant 0 Confused

You didn't said how MUCH of A 1 A_1 we have. So there is no correct answer.

In your solution, you assumed to have 138 g of C2H6O ≈ \approx 3 mol.

I think this is clear to you, but molality is an intensive property, saying 3 mol kg − 1 3 \text{ mol kg}^{-1} does not tell us about the initial mass.

Bernardo Sulzbach - 6 years, 11 months ago

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