18g glucose is dissolved in 90g of water. The relative lowering of vapour pressure is equal to?
Answer: 1/50
Explanation
The relative lowering of vapor pressure is given by the formula:
Relative lowering of vapor pressure = (Mole fraction of solute)/(Mole fraction of solvent + Mole fraction of solute)
Calculating the mole fractions:
Moles of glucose = (18 g / 180 g/mol) = 0.1 mol
Moles of water = (90 g / 18 g/mol) = 5 mol
Mole fraction of glucose = 0.1 / (0.1 + 5) = 0.02
Mole fraction of water = 1 - 0.02 = 0.98
Substituting into the formula:
Relative lowering of vapor pressure = 0.02 / (0.02 + 0.98) = 0.02 / 1 = 0.02
Therefore, the relative lowering of vapor pressure is 0.02, which is equivalent to 1/50.
