A) 91.8 g
B) 1198 g
C) 83.9 g
D) 890.3g
Correct Answer: A
Solution :
\[{{\pi }_{glycerine}}={{\pi }_{glu\cos e}}\] \[\frac{{{n}_{1}}}{{{V}_{1}}}RT=\frac{{{n}_{2}}}{{{V}_{2}}}RT\] \[\frac{10.2}{M}\times \frac{1}{1}=\frac{2}{180}\times \frac{1000}{100}\] \[\Rightarrow M=\frac{10.2\times 18}{2}=91.8g\] (Density of water =\[1g/c{{m}^{3}}\])You need to login to perform this action.
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