• question_answer 12 g of a non-volatile solute dissolved in 108 g of water produces the relative lowering of vapour pressure of 0.1. The molecular mass of the solute is A)  80                B)  60         C)  20                 D)  40

From Raoults law, relative lowering in vapour pressure $\Delta p=\frac{{{p}^{o}}-p}{{{p}^{o}}}=\frac{n}{N}=\frac{W}{m}\times \frac{M}{W}$ Here, $w=12g;W=108g,.m=?$ $M=18g,\Delta p=0.1$ $\Delta p=\frac{W}{m}\times \frac{M}{W}$ or $0.1=\frac{12}{m}\times \frac{18}{108}$ $m=\frac{12\times 18}{10.8}=20$