$\begin{array}{1 1}(a)\;0.0372K\\(b)\;0.112K\\(c)\;0.0744K\\(d)\;0.0186K\end{array}$

Want to ask us a question? Click here

Browse Questions

Ad |

0 votes

0 votes

Answer: 0.1120$^{\circ}$K

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.

Calculating $i$ for Sodium Sulfate $\rightarrow Na_2SO_4^{2-} (s) \rightarrow 2Na^{+} + SO_4^{2-} \rightarrow i = 3$

In this case $i = 3, \; K_f = 1.86$ and $m = 0.02 \rightarrow \Delta T_f = 3 \times 1.86 \times 0.02 = 0.112$

Ask Question

Take Test

x

JEE MAIN, CBSE, AIPMT Mobile and Tablet App

The ultimate mobile app to help you crack your examinations

...