$\begin{array}{1 1}(a)\;-5.7\times 10^{-2}\\(b)\;-5.7\times 10^{-3}\\(c)\;-1.2\times 10^{-2}\\(d)\;-2.3\times 10^{-2}\end{array}$

Want to ask us a question? Click here

Browse Questions

Ad |

0 votes

0 votes

Answer: -0.0226 $^{\circ}$C

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.

$\Rightarrow$ Molarity $= \large\frac{W}{M} $$ \times\large\frac {1000}{V\;mL}$$ = \large\frac{0.1}{329}$$\times \large\frac{100}{1000}$$ =0.003039$

$\Rightarrow \Delta T_f = 4 \times 1.86 \times 0.003039 = 0.02261$ which is equal to $T_0 - T_f$

Since, $T_0 = 273^{\circ}K = 0^{\circ}C \rightarrow T_f = -0.0226^{\circ}C$

Ask Question

Take Test

x

JEE MAIN, CBSE, AIPMT Mobile and Tablet App

The ultimate mobile app to help you crack your examinations

...