Glucose is added in 100 gm of water. Lowering in vapor pressure is 0.2 mm Hg. Vapour pressure of pure water is 54.2 mm Hg. Then the weight of glucose is?
3.70 gm
4.92 gm
6.73 gm
8.74 gm
The correct answer is option (A): 3.70 gm
\(\frac{0.2}{54}=\frac{n_{glucose}}{\frac{100}{18}}\)
\(n_{glucose}\frac{0.2}{54}=\frac{100}{18}\)
Mass of glucose = \(\frac{0.2}{54}\times\frac{100}{18}\times180=3.70\,gm\)