A) 14
B) 3.2
C) 1.4
D) 2
Correct Answer: D
Solution :
[d] 3.2 \[{{x}_{2}}=\] amount of solute \[=\frac{{{n}_{2}}}{{{n}_{1}}+{{n}_{2}}}=0.2\] \[\therefore \,{{x}_{1}}=1-0.2=0.8\] \[=\frac{{{n}_{2}}}{{{n}_{1}}+{{n}_{2}}}=0.8\] \[\therefore \frac{{{x}_{2}}}{{{x}_{1}}}=\frac{{{n}_{2}}}{{{n}_{1}}}=\frac{1}{4}\] Taking \[{{n}_{1}}1000/78\]moles (mol. mass of\[{{C}_{6}}{{H}_{6}}=78\]) \[{{n}_{2}}=\frac{1}{4}\times {{n}_{1}}\] \[=\frac{1}{4}\times \frac{1000}{78}=3.20\]
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