A) 30 g
B) 60 g
C) 120 g
D) 12 g
E) 24 g
Correct Answer: B
Solution :
Relative lowering of vapour pressure is given by the formula \[\frac{p{}^\circ -{{p}_{s}}}{p{}^\circ }=\frac{{{w}_{A}}}{{{M}_{A}}}\times \frac{{{M}_{B}}}{{{w}_{B}}}\] As vapour pressure of water is lowered by 10%. \[\therefore \] \[\frac{p{}^\circ -{{p}_{s}}}{p{}^\circ }=\frac{10}{100}\] \[\therefore \] \[\frac{10}{100}=\frac{{{w}_{A}}}{60}\times \frac{18}{180}\] or \[{{w}_{A}}=60\text{ }g\]You need to login to perform this action.
You will be redirected in
3 sec