Question

What will be the boiling point of water if $1$ mole of NaCl is dissolved in $1,000$ grams of water?

• A. ${100.51}^o$C
• B. ${101.02}^o$C
• C. ${101.53}^o$C
• D. ${101.86}^o$C
• E. ${103.62}^o$C

Answer 1

Answer: (B)

${101.02}^o$C

1 mole of NaCl is dissolved in 1,000 grams of water.
The molality, m, of NaCl is the ratio of number of moles of NaCl to mass of water (in kg)$=\frac{1mol}{1000g\times \frac{1kg}{1000g}}=1m$.
The elevation in the boiling point $\Delta T_b=i{K}_{b}m=2\times 0.51={1.02}^{o}C$.
(Here i 2 is the Vant Hoff's factor. 1 molecule of NaCl on dissociation gives 2 ions. Hence, i=2 and Kb is the molal elevation in the boiling point constant)
The boiling point of pure water is ${100}^{o}C$.
The boiling point of aq NaCl solution is $100+1.02={101.02}^{o}C$.