Step 1: Let total moles = 1 mole. Given:
- Mole fraction of glucose = 0.0244
- Mole fraction of water = 0.9756 So, moles of glucose = 0.0244 mol, and moles of water = 0.9756 mol.
Step 2: Use molar masses.
- Molar mass of glucose (C$_6$H$_{12}$O$_6$) = 180 g/mol
- Molar mass of water (H$_2$O) = 18 g/mol Mass of glucose = \( 0.0244 \times 180 = 4.392 \) g
Mass of water = \( 0.9756 \times 18 = 17.5608 \) g
Step 3: Calculate weight percentage of glucose. \[ w/w % of glucose} = \frac{4.392}{4.392 + 17.5608} \times 100 \approx \frac{4.392}{21.9528} \times 100 \approx 20% \]
A substance 'X' (1.5 g) dissolved in 150 g of a solvent 'Y' (molar mass = 300 g mol$^{-1}$) led to an elevation of the boiling point by 0.5 K. The relative lowering in the vapour pressure of the solvent 'Y' is $____________ \(\times 10^{-2}\). (nearest integer)
[Given : $K_{b}$ of the solvent = 5.0 K kg mol$^{-1}$]
Assume the solution to be dilute and no association or dissociation of X takes place in solution.