A) \[0.2\]
B) \[0.4\]
C) \[0.6\]
D) \[0.8\]
Correct Answer: B
Solution :
On applying this relation \[\frac{{{p}^{o}}-{{p}_{s}}}{{{p}^{o}}}=\frac{n}{n+N}\] Where, \[{{p}^{o}}-{{p}_{s}}=\] lowering of vapour pressure \[{{p}^{o}}=\] vapour pressure of pure solvent \[\frac{n}{n+N}\] mole fraction of solute. \[\Rightarrow \] \[\frac{10}{{{p}^{o}}}=0.2\] \[\therefore \] \[{{p}^{o}}=\frac{10}{0.2}\] ?..(i) Again, \[\frac{{{p}^{o}}-{{p}_{s}}}{{{p}^{o}}}=\frac{n}{n+N}\] \[\Rightarrow \] \[\frac{20}{{{p}^{o}}}={{X}_{n}}\][Mole fraction of solute] \[\therefore \] \[{{p}^{o}}=\frac{20}{{{X}_{n}}}\] ?.(ii) From Eqs. (i) and (ii) \[\Rightarrow \] \[\frac{10}{0.2}=\frac{20}{{{X}_{n}}}\] \[\therefore \] \[{{X}_{n}}=0.4\] Mole fraction of solute \[=0.4\]
You need to login to perform this action.
You will be redirected in
3 sec
