1. Tardigrade
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  4. If 1 mole of NaCl solute is dissolved into the 1 kg of water, at what temperature will water boil at 1.013 bar? (Kb of water is 0.52 K kg mol-1)

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Q. If $1\, mole$ of NaCl solute is dissolved into the $1\, kg$ of water, at what temperature will water boil at 1.013 bar? ($K_b$ of water is $0.52 \, K \, kg \, mol^{-1}$)

KEAMKEAM 2018Solutions

Solution:

$\because \Delta T= K_{b} \frac{n}{w_{A}}$

Given, $K_{b} =0.52 K kg mol ^{-1}$

$n =1, w_{A}=1\, kg$

$\therefore \Delta T =\frac{0.52 \times 1}{1}=0.52\, K$

$\because$ Boiling point of pure water at $1.013$ bar (i.e. $1 atm$ ) is $373.15\, K$

$\therefore $ Boiling point of the solulion $=373.15+0.52$

$=373.67\, K$