A) 80
B) 60
C) 20
D) 40
Correct Answer: C
Solution :
From Raoult's law Relative lowering in vapour pressure \[\Delta p=\frac{{{p}^{0}}-p}{{{p}^{0}}}=\frac{n}{N}=\frac{w}{m}\times \frac{M}{W}\] \[w=12\,g;\,\,W=108\,g,\] \[m=?;\] \[M=18\,g,\,\,\Delta p=0.1\] \[\Delta p=\frac{w}{m}\times \frac{M}{W}\] \[0.1=\frac{12}{m}\times \frac{18}{108}\] \[m=\frac{12\times 18}{10.8}=20\]You need to login to perform this action.
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