Question

Change in entropy $\left(\text{in J/K}\right)$ for freezing of $10\text{g}$ of $H_{2}O\left(l\right)$ (enthalpy of fusion is $80\text{cal/g}$) at $0^{\circ }\text{C}$ and $1\text{atm}$ is:
(Given: $1\text{cal}=4.2\text{J}$, Round the answer upto one decimal point)

• A. $24.6\text{J/K}$
• B. $12.3\text{J/K}$
• C. $18.5\text{J/K}$
• D. $9.2\text{J/K}$

Answer 1

The correct option is B $12.3\text{J/K}$

The change in entropy is $\Delta S_{\text{freezing}}=-\frac{\Delta H_{\text{fusion}}}{T}=-\frac{80\times 10}{273}=-2.93\text{cal/K}=12.3\text{J/K}$