Question:
For a dilute solution containing 2.5 g of a non-volatile non-electrolyte solute in 100g of water, the elevation in boiling point at 1 atm pressure is 2$^{\circ}$C. Assuming
concentration of solute is much lower than the concentration of solvent, the vapour pressure (mm of Hg) of the solution is (take $K_b = 0.76 K kg mol^{-1}).$
For a dilute solution containing 2.5 g of a non-volatile non-electrolyte solute in 100g of water, the elevation in boiling point at 1 atm pressure is 2$^{\circ}$C. Assuming
concentration of solute is much lower than the concentration of solvent, the vapour pressure (mm of Hg) of the solution is (take $K_b = 0.76 K kg mol^{-1}).$
Updated On: Jun 23, 2023
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The Correct Option is A
Solution and Explanation
The elevation in boiling point is
$\Delta T_b=K_b.m:m=molality=\frac{n_2}{w_1} \times 1000$
$ [n_2 =Number \, of \, moles \, of \, solute , \, w_1 =$ Weight of
$\hspace50mm $ solvent in gram]
$\Rightarrow \, 2=0.76 \times \frac{n_2}{100} \times 1000$
$\Rightarrow n_2 =\frac{5}{19}$
Also, from Raoult- law of lowering of vapour pressure :
$ \frac{-\Delta p}{p^{\circ}}=x_2 =\frac{n_2}{n_1+n_2}=\frac{n_2}{n_1} \, \, \, \, [\because \, n_1 > > n_2]$
$\Rightarrow =\Delta p =760 \times \frac{5}{19} \times \frac{18}{100}$
$\hspace30mm =36 \, mm \, of \, Hg$
$\Rightarrow p=760-36=724 \, mm \, of \, Hg$
$\Delta T_b=K_b.m:m=molality=\frac{n_2}{w_1} \times 1000$
$ [n_2 =Number \, of \, moles \, of \, solute , \, w_1 =$ Weight of
$\hspace50mm $ solvent in gram]
$\Rightarrow \, 2=0.76 \times \frac{n_2}{100} \times 1000$
$\Rightarrow n_2 =\frac{5}{19}$
Also, from Raoult- law of lowering of vapour pressure :
$ \frac{-\Delta p}{p^{\circ}}=x_2 =\frac{n_2}{n_1+n_2}=\frac{n_2}{n_1} \, \, \, \, [\because \, n_1 > > n_2]$
$\Rightarrow =\Delta p =760 \times \frac{5}{19} \times \frac{18}{100}$
$\hspace30mm =36 \, mm \, of \, Hg$
$\Rightarrow p=760-36=724 \, mm \, of \, Hg$
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Concepts Used:
Solutions
A solution is a homogeneous mixture of two or more components in which the particle size is smaller than 1 nm.
For example, salt and sugar is a good illustration of a solution. A solution can be categorized into several components.
Types of Solutions:
The solutions can be classified into three types:
- Solid Solutions - In these solutions, the solvent is in a Solid-state.
- Liquid Solutions- In these solutions, the solvent is in a Liquid state.
- Gaseous Solutions - In these solutions, the solvent is in a Gaseous state.
On the basis of the amount of solute dissolved in a solvent, solutions are divided into the following types:
- Unsaturated Solution- A solution in which more solute can be dissolved without raising the temperature of the solution is known as an unsaturated solution.
- Saturated Solution- A solution in which no solute can be dissolved after reaching a certain amount of temperature is known as an unsaturated saturated solution.
- Supersaturated Solution- A solution that contains more solute than the maximum amount at a certain temperature is known as a supersaturated solution.