The entropy of vaporization of benzene is $85{\mathrm{JK}}^{-1} {\mathrm{mol}}^{-1}$. When $117 \mathrm{g}$ benzene vaporizes at it's normal boiling point, the entropy change of surrounding is :

Benzene's vaporization's entropy $\mathrm{\Delta s}=85{\mathrm{Jk}}^{-1}{\mathrm{mol}}^{-1}$

Benzene content = 117 g

Molar mass of Benzene $\left({\mathrm{C}}_6{\mathrm{H}}_6\right)=78\mathrm{g}$

Number of moles$=117/78=1.5\text{ mole }$

$\mathrm{\Delta }{S}_{\mathrm{sy}8}=85\times \frac{117}{78}=-\mathrm{\Delta }{S}_{\text{surr }}$

Entropy in the immediate environment$=-85\times 1.5$