Question

A beaker containing ${\text{20 g}}$ sugar in ${\text{100 g}}$ water and another containing ${\text{10 g}}$ sugar in ${\text{100 g}}$ water, are placed under a bell jar and allowed to stand until equilibrium is reached. Amount of water will be transferred from one beaker to another in a decagram (nearest integer).

Answer 1

To answer this question, you should recall the concept of molarity. Molarity is defined as the moles of a solute per litre of a solution. We shall substitute the values in the given equation and compare it with each other.

Formula used:
${\text{Molarity = }}\dfrac{{{\text{no}}{\text{. of moles}}}}{{{\text{Volume}}}}$

Complete step by step solution:
From the formula, we can calculate the molarity of each of the solutions and equate the concentrations of both the solutions. As density of water $1{\text{g/ml}}$ so the volume of ${\text{100 g}}$ water will be ${\text{100 ml }}$.
Molarity of ${\text{10 g}}$ sugar in ${\text{100 ml }}$ water$= \left( {\dfrac{{10}}{{342}}} \right)\left( {\dfrac{{1000}}{{100}}} \right) = 0.2932{\text{M}}$.
Now, Molarity of $20{\text{g}}$ sugar in ${\text{100 ml }}$ water $= \left( {\dfrac{{20}}{{342}}} \right)\left( {\dfrac{{1000}}{{100}}} \right) = 0.5840{\text{M}}$.
We can now equate the values of the concentrations.
$\therefore$$${{\text{M}}{\text{1}}}{{\text{V}}{\text{1}}} = {{\text{M}}{\text{2}}}{{\text{V}}{\text{2}}}$$.

So, to make equal concentration equal amount of $= 33.3{\text{g}} \approx 3{\text{dag}}$ must be transferred out to the other beaker.

Note:
You should know about the other concentration terms commonly used:

1. Concentration in Parts Per Million (ppm) The parts of a component per million parts (${10^6}$) of the solution.
2. ${\text{ppm(A)}} = \dfrac{{{\text{Mass of A}}}}{{{\text{Total mass of the solution}}}} \times {10^6}$
3. Molality (m): Molality establishes a relationship between moles of solute and the mass of solvent. It is given by moles of solute dissolved per kg of the solvent. The molality formula is as given- ${\text{Molality(m) = }}\dfrac{{{\text{Moles of solute}}}}{{{\text{Mass of solvent in kg}}}}$
4. Normality: It is defined as the number of gram equivalents of solute present in one litre of the solution.
5. Mole fraction: It gives a unitless value and is defined as the ratio of moles of one component to the total moles present in the solution. Mole fraction = $\dfrac{{{{\text{X}}_{\text{A}}}}}{{{{\text{X}}_{\text{A}}} + {{\text{X}}_{\text{B}}}}}$(from the above definition) where ${{\text{X}}_{\text{A}}}$ is no. of moles of glucose and ${{\text{X}}_{\text{B}}}$ is the no. of moles of solvent