1. Tardigrade
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  3. Chemistry
  4. The vapour pressure of water is 12.3 kPa at 300 K. What is the vapour pressure of 1 molal aqueous solution of a non-volatile solute at 300 K ?

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Q. The vapour pressure of water is $12.3\, kPa$ at $300\, K$. What is the vapour pressure of 1 molal aqueous solution of a non-volatile solute at $300 \,K ?$

Solutions

Solution:

$\frac{ P ^{0}- P _{ S }}{ P _{ S }}=\frac{ n _{ B }}{ n _{ A }}$

1 molal means 1 mole of solute is dissolved in $1 kg$ of solvent (water here)

$\frac{12.3- P _{ S }}{12.3}=\frac{1}{1+\frac{1000}{18}}=\frac{1}{1+55.56}$

$\frac{12.3- P _{ S }}{12.3}=0.0177$

$12.3- P _{ s }=0.2177$

$P _{ s }=12.08 kPa$