Question

Which of the following solutions will exhibit highest boiling point?

• A. 0.01 M $N{a}_{2}S{O}_4$
• B. 0.01 M $KN{O}_3$
• C. 0.015 M urea
• D. 0.015 M glucose

Answer 1

The correct option is A 0.01 M $N{a}_{2}S{O}_4$

$△=i\times k_b\times m$
$i\times m$ of $N{a}_{2}S{O}_4$ is highest, hence its boiling point will also be highest.
$N{a}_{2}S{O}_4$ $i\times m=3\times 0.01=0.03$
$KN{O}_3$ $i\times m=2\times 0.01=0.02$
Urea $i\times m=1\times 0.015=0.015$
Glucose $i\times m=1\times 0.015=0.015$