Question

Which one of the following aqueous solutions will exhibit highest elivation in boiling point?

• A. 0.05 M glucose (non ionisable)
• B. 0.01 M $KNO_{3}$ (50 % ionisable)
• C. 0.015 M Urea (non ionisable)
• D. 0.01 M $Na_{2}SO_{4}$ (75 % ionisable)

Answer 1

Answer: (A)

Boiling point

$=T_{0}\left(s o l v e n t\right)+\Delta T_{b}\left(e l e v a t i o n \, i n \, b . p .\right)$

$\Delta T_{b}=miK_{b}$

where, m is the molality $\left(\approx m o l a r i t y\right)$

If n is number of species formed after dissociation

i, the vant Hoff factor $=\left[1 + \left(n - 1\right) \alpha \right]$

$K_{b}=$ molal elevation constant.

Thus $\Delta T_{b} \propto i$

Assume 100% ionization

$\left[1\right]m \, i \, \left(N a_{2} S O_{4}\right)=0.01\left[1 + \left(3 - 1\right) 0.75\right]=0.025$

$\left[2\right]m \, i \, \left(K N O_{3}\right)=0.01\times \left[1 + \left(2 - 1\right) 0.50\right]=0.015$

$\left[3\right]m \, i \, \left(u r e a\right)=0.015$

$\left[4\right]m \, i \, \left(g l u c o s e\right)=0.05$