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 = \(\frac{0.2}{54}\times\frac{100}{18}\times180=3.70\,gm\)
Observe the following data given in the table. (\(K_H\) = Henry's law constant)
| Gas | CO₂ | Ar | HCHO | CH₄ |
|---|---|---|---|---|
| \(K_H\) (k bar at 298 K) | 1.67 | 40.3 | \(1.83 \times 10^{-5}\) | 0.413 |
The correct order of their solubility in water is
A square loop of sides \( a = 1 \, {m} \) is held normally in front of a point charge \( q = 1 \, {C} \). The flux of the electric field through the shaded region is \( \frac{5}{p} \times \frac{1}{\varepsilon_0} \, {Nm}^2/{C} \), where the value of \( p \) is: