Question

Lowering of vapour pressures of equimolar solution of glucose, sodium chloride and barium nitrate are in the order:

• A. Glucose > $NaCl$ > $Ba(N{O}_3{)}_2$
• B. Glucose = $NaCl$ = $Ba(N{O}_3{)}_2$
• C. $Ba(N{O}_3{)}_2$ > $NaCl$ > Glucose
• D. $NaCl$ > $Ba(N{O}_3{)}_2$ > Glucose

Answer 1

The correct option is C $Ba(N{O}_3{)}_2$ > $NaCl$ > Glucose

Van't Hoff factor $=i$

For $Ba(N{O}_3{)}_2,i=3;$ $Ba+2N{O}_3^{-}$

For $NaCl$, $i=2;$ $N{a}^{+}+C{l}^{-}$

For Glucose, $i=1$

So higher the van't Hoff factor, higher is the lowering of vapor pressure

So $Ba(N{O}_3{)}_2$ will have the highest lowering of vapor pressure.

Option C is correct.