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Question: The vapour pressure of a distilled water at \(298K\) is \(5.0 \times {10^4}N{m^{ - 2}}\) . When a no...

The vapour pressure of a distilled water at 298K298K is 5.0×104Nm−25.0 \times {10^4}N{m^{ - 2}} . When a non-volatile solute is dissolved, the vapour pressure becomes 4.25×104Nm−24.25 \times {10^4}N{m^{ - 2}} . Calculate relative lowering of vapour pressure and lowering of vapour pressure .

Explanation

Solution

The vapour pressure is a measure of the material's tendency to switch from gaseous or vapour to an increasing temperature. The temperature at which a vapour pressure on the surface of a fluid equals the surrounding pressure is known as the boiling point.

Complete answer: P0{P_0} = Vapour pressure of solvent
P0{P_0} = 5.0×104Nm−25.0 \times {10^4}N{m^{ - 2}}
PS{P_S} = Vapour pressure of solution
PS{P_S} = 4.25×104Nm−24.25 \times {10^4}N{m^{ - 2}}
A solution containing a non-volatile solvent has less steam pressure at any given temperature than a pure solvent. This is known as the lowering of vapour pressure.
Non-volatile means a small tendency to evaporate the solution itself. Because some of the surface now contains solvent particles, solvent molecules have less room. This leads to the ability to evaporate less solvent. Adding a non-volatile solution results in a decrease in the solvent vapour pressure.
Lowering in vapour pressure:
=P0−PS =5×104−4.25×104 =0.75×104Nm−2 =7.5×103Nm−2  = {P_0} - {P_S} \\\ = 5 \times {10^4} - 4.25 \times {10^4} \\\ = 0.75 \times {10^4}N{m^{ - 2}} \\\ = 7.5 \times {10^3}N{m^{ - 2}} \\\
When a non-volatile solution has been incorporated into a volatile solvent, the vapour pressure is lower than the vapour pressure of pure solvent. It is due to an increase in density that reduces the evaporation rate.
Relative lowering in vapour pressure = P0−P5P0\dfrac{{{P_0} - {P_5}}}{{{P_0}}}
⇒P0−P5P0=7.5×1035×104=1.5×10−1  ⇒P0−P5P0=0.15  \Rightarrow \dfrac{{{P_0} - {P_5}}}{{{P_0}}} = \dfrac{{7.5 \times {{10}^3}}}{{5 \times {{10}^4}}} = 1.5 \times {10^{ - 1}}\, \\\ \Rightarrow \dfrac{{{P_0} - {P_5}}}{{{P_0}}} = 0.15 \\\
Hence, the lowering in vapour pressure is .5×103Nm−2.5 \times {10^3}N{m^{ - 2}} and relative lowering in vapour pressure is 0.150.15 .

Note:
The vapour pressure decrease depends on the amount of dissolved solute particles. The solution's chemical nature is not important because the vapour pressure is simply a physical feature of the solvent. The only requirement is that the solvent is dissolved, and that the solvent does not suffer a chemical reaction.