A) 74.2
B) 75.6
C) 67.83
D) 78.7
Correct Answer: C
Solution :
We have \[\frac{{{p}^{0}}-21.85}{21.85}=\frac{30\times 18}{90\times m},\,\,\] for I case (i) Weight of solvent \[= 90 + 18 = 108 g\] \[\frac{{{p}^{0}}-22.15}{22.15}=\frac{30\times 18}{108\times m},\] for II case (ii) By eq.[A] \[{{p}^{0}}_{m}\,=\,\,21.85m\,\,=\,\,21.85\,\,\times \,\,6=131.1\] By eq. [B] \[p_{m}^{0}-22.15\,m\, =\, 22.15 \times 5 = 110.75\] \[0.30\,m\,\,=\,\,20.35\] \[m=\frac{20.35}{0.30}=67.83\]You need to login to perform this action.
You will be redirected in
3 sec