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

Hint:
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