Chemistry Daily Challenge 16-July-2015

Chemistry Level 2

A beaker containing 20 g of sugar in 100 g of water and another containing 10 g of sugar in 100 g of water are placed under a bell-jar and allowed to stand until equilibrium is reached. How much water (in grams) will be transferred from one beaker to another? Round your answer to two decimal points.

Note: Ignore the water that stays vaporized.

The answer is 33.33.

1 solution

Chew-Seong Cheong
Jul 24, 2015

Under a vacuum condition water will transfer from solution of higher concentration to the lower one until the two concentrations equalize. Let the amount of water transferred be $m$ then, we have:

$\begin{aligned} \frac{20}{100+m} & = \frac{10}{100-m} \\ \frac {100+m}{20} & = \frac {100-m}{10} \\ 100 + m & = 200 - 2m \\ 3m & = 100 \\ m & = \boxed{33.33} \end{aligned}$

Moderator note:

A slightly easier (but equivalent) approach would be to observe that the amount of sugar is 2:1, hence the amount of water in equilibrium should also be 2:1.

That's a perfect solution sir!

Aamir Faisal Ansari - 5 years, 10 months ago

Thanks. Just doing the best I can.

Chew-Seong Cheong - 5 years, 10 months ago

Does concentration here stand for concentration of water instead of concentration of sugar?

Lu Chee Ket - 5 years, 4 months ago

Concentration here means grams of sugar in a litre of water. More sugar more concentrated solution. Less sugar or more water less concentrated or more diluted solution.

Chew-Seong Cheong - 5 years, 4 months ago

Shouldn't water transfer in opposite direction to what you described then?