A) 20 g
B) 10 g
C) 30 g
D) 45 g
Correct Answer: A
Solution :
[b]| According to Raoul?s law, relative lowering of vapour pressure, \[\frac{p_{A}^{{}^\circ }-ps}{p_{A}^{{}^\circ }}={{\chi }_{B}}\] |
| \[{{\chi }_{B}}=\frac{{{n}_{B}}}{{{n}_{B}}+{{n}_{A}}}=\frac{{{W}_{B}}/{{M}_{B}}}{\frac{{{W}_{B}}}{{{M}_{B}}}+\frac{{{W}_{A}}}{{{M}_{A}}}}\] |
| Given, vapour pressure is reduced to 80 when non-volatile solute is dissolved in octane. It means |
| If \[p_{A}^{{}^\circ }=1\,\text{atm}\]then \[{{p}_{S}}=0.8\,\text{atm};\] \[p_{A}^{{}^\circ }-{{p}_{S}}=0.2\,\text{atm};\] \[{{M}_{A}}({{C}_{8}}{{H}_{18}})=114\,g\,\text{mo}{{\text{l}}^{-\,1}};\,\,{{W}_{A}}=114g;\] |
| \[{{M}_{B}}=40g\,\text{mo}{{\text{l}}^{-\,1}};\,\,{{W}_{B}}=?\] |
| \[{{M}_{B}}=40g\,mo{{l}^{-1}};\,{{W}_{B}}=?\] |
| On applying Eq. (i) and (ii), we get, \[\frac{0.2}{1}=\frac{{{W}_{B}}40}{\frac{{{W}_{B}}}{40}+\frac{114}{114}}=\frac{{{W}_{B}}/40}{\frac{{{W}_{B}}}{40}+1}\] |
| \[0.2=\frac{{{W}_{B}}}{{{W}_{B}}+40}\] |
| \[0.2{{W}_{B}}+8={{W}_{B}}\] |
| \[{{W}_{B}}=10g\] |
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