Question:
The vapour pressure of water is 12.3 kPa at 300 K. Calculate vapour pressure of 1 molal solution of a non-volatile solute in it
The vapour pressure of water is 12.3 kPa at 300 K. Calculate vapour pressure of 1 molal solution of a non-volatile solute in it
Updated On: Sep 27, 2023
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Solution and Explanation
The correct answer is: \(12.08 kPa\)
1 molal solution means 1 mol of the solute is present in 1000 g of the solvent (water).
Molar mass of water \(= 18 g mol^{- 1}\)
Number of moles present in 1000 g of water\(= \frac{1000}{18}\)
\(= 55.56\, mol\)
Therefore, mole fraction of the solute in the solution is
\(x_2 = \frac{1}{(1+55.56)}= 0.0177\)
It is given that,
Vapour pressure of water,\(p^0_1 = 12.3 kPa \)
Applying the relation,\(\frac{(p^o_1-p_1)}{p^o_1} = x_2\)
\(⇒\frac{(12.3-p_1)}{12.3}= 0.0177\)
\(⇒ 12.3 - p_1 = 0.2177\)
\(⇒ p_1 = 12.0823 \)
\(= 12.08 kPa\) (approximately)
Hence, the vapour pressure of the solution is \(12.08 kPa\)
1 molal solution means 1 mol of the solute is present in 1000 g of the solvent (water).
Molar mass of water \(= 18 g mol^{- 1}\)
Number of moles present in 1000 g of water\(= \frac{1000}{18}\)
\(= 55.56\, mol\)
Therefore, mole fraction of the solute in the solution is
\(x_2 = \frac{1}{(1+55.56)}= 0.0177\)
It is given that,
Vapour pressure of water,\(p^0_1 = 12.3 kPa \)
Applying the relation,\(\frac{(p^o_1-p_1)}{p^o_1} = x_2\)
\(⇒\frac{(12.3-p_1)}{12.3}= 0.0177\)
\(⇒ 12.3 - p_1 = 0.2177\)
\(⇒ p_1 = 12.0823 \)
\(= 12.08 kPa\) (approximately)
Hence, the vapour pressure of the solution is \(12.08 kPa\)
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