$\begin{array}{1 1}(a)\;\text{1% of solution of urea in water}\\(b)\;\text{1% solution of zinc sulphate in water}\\(c)\;\text{1% solution magnesium chloride in water}\\(d)\;\text{1% solution of glucose in water}\end{array}$

Want to ask us a question? Click here

Browse Questions

Ad |

0 votes

0 votes

Answer: - 1% solution of $MgCl_2$

Depression of Freezing point, $\Delta\; T_f = i\;K_f\; m$, where $i$ is the van't Hoff factor, $K_f$ is the Freezing Point Depression Constant and $m$ is the molality.

Molality $m = \large\frac{w/M}{W(g)/1000}$

For the given problem, the solutions are all 1% solution and hence Molality $m \propto \large\frac{1}{M}$

Therefore $\Delta T_f \propto \large\frac{i}{M}$

Note: Molecular mass of Glucose $= 180\;g\;mol^{-1}$, Urea $ = 60\;g\;mol^{-1}$, $MgCl_2 = 95.2\;g\;mol^{-1}$, $ZnSO_4 = 161.5\;g\;mol^{-1}$,

Note: van't Hoff factor for Glucose $ = 1$, Urea $ = 1$, $ZnSO_4 = 2$ and $MgCl_2 = 3$.

$\Rightarrow$ the solution with the lowest freezing point based on the above data is $MgCl_2$.

Ask Question

Take Test

x

JEE MAIN, CBSE, NEET Mobile and Tablet App

The ultimate mobile app to help you crack your examinations

...