Question

Arrange the following aqueous solutions in the order of their increasing boiling points:
(i) ${10}^{-4}M$ $NaCl$
(ii) ${10}^{-4}M$ Urea
(iii)${10}^{-3}M$ $Mg{Cl}_2$
(iv) ${10}^{-2}M$ $NaCl$

• A. $\left(i\right)<\left(ii\right)<\left(iv\right)<\left(iii\right)$
• B. $\left(ii\right)<\left(i\right)<\left(iii\right)<\left(iv\right)$
• C. $\left(iii\right)<\left(i\right)<\left(ii\right)<\left(iv\right)$
• D. $\left(iv\right)<\left(iii\right)<\left(i\right)=\left(ii\right)$

Answer 1

The correct option is B $\left(ii\right)<\left(i\right)<\left(iii\right)<\left(iv\right)$

Given concentration $i$ $i\times concentration$

i) ${10}^{-4}M$ $NaCl$ $2$ $2\times {10}^{-4}M$

ii) ${10}^{-4}M$ urea $1$ $1\times {10}^{-4}M$

iii) ${10}^{-3}M$ ${MgCl}_2$ $3$ $3\times {10}^{-3}M$

iv) ${10}^{-2}M$ $NaCl$ $2$ $2\times {10}^{-2}M$

As concentration is increased $B.P$. of the solution also increases.

Order of Boiling Point $(B.P)$ will be same as the order of increasing concentration:-

$ii>i>iii>iv$