Question

what is the boiling point of 1 molal aqueous solution of NaCl (kb=0.52k/molal)

Answer 1

Answer:

The boiling point of 1 molal aqueous solution of NaCl is $\bold{101.04^{\circ} \mathrm{C}}$

Explanation:

Given:  

Molality of solution = 1 m

The Vant Hoff factor for NaCl = 2 ( as it can form two ions)  

$Kb = 0.52\ K/molal$

$\Delta \mathrm{Tb}=i \times K b \times molality$

Plug the given values:

$\Delta \mathrm{Tb}=2 \times 0.52\ \mathrm{K} / \mathrm{mol} \times 1\ \mathrm{molal}$

$\Delta \mathrm{Tb}=1.04$

Boiling point of water $=100^{\circ} \mathrm{C}$

Hence boiling point of solution $=100+1.04=\bold{101.04^{\circ} \mathrm{C}}$

Answer 2

Explanation:

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