A) 128
B) 488
C) 32
D) 64
Correct Answer: D
Solution :
vapour pressure of Acetone = 185 torr molar mass of Acetone\[(C{{H}_{3}}COC{{H}_{3}})=58\] \[{{n}_{acetone}}=\frac{100}{58}=1.724\] \[{{n}_{subs\tan ce}}=\frac{1.2}{M}\] by\[\frac{{{P}^{0}}-{{P}_{s}}}{{{P}^{0}}}=\frac{{{n}_{subs\tan ce}}}{{{n}_{subs\tan ce}}+{{n}_{acetone}}}\] \[\frac{185-183}{185}=\frac{\frac{1.2}{M}}{\frac{1.2}{M}+1.724}\] \[\frac{2}{185}=\frac{\frac{1.2}{M}}{\frac{1.2}{M}+1.724}\] \[\frac{185}{2}=1+\frac{1.724}{1.2/M}\] \[\frac{185-2}{2}=\frac{1.724}{1.2}M\] \[M=63.68\approx 64\]
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