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) $10^{-3}\,M\,MgCl_2$
(iv) $10^{-2}\, M\, NaCl$

• A. (i) < (ii) < (iv) < (iii)
• B. (ii) < (i) = (iii) < (iv)
• C. (ii) < (i) < (iii) < (iv)
• D. (iv) < (iii) < (i) = (ii)

Answer 1

Answer: (C)

For $10^{-4}\, M \,NaCl\, i = 2$
$10^{-4}\, M$ Urea $i = 1$
$10^{-3}\,M\,MgCl_2\, i = 3$
$10^{-2}\, M\, NaCl \, i = 2$
More the value of i, C, more will be the elevation in boiling point hence increasing order of boding point is $10^{-4}\,M$ Urea $< 10^{-4}\, M\, NaCl < 10^{-3}\, M\, MgCl_2 < 10^{-2}\, M\, NaCl$