A solution is obtained by dissolving 6 g of urea (mol. wt. = 60) in a litre solution, another solution is prepared by dissolving 34.2 g of cane sugar (mol. wt. = 342) in a litre of solution at the same temperature The lowering of vapour pressure in the first solution is:
Answer & explanation
Correct answer: option 1
The correct answer is option 1. Same as that of second solution.
The lowering of vapor pressure in a solution is given by the equation:
\(\Delta P = i \cdot X \cdot P^o \)
Where:
\( \Delta P \) is the lowering of vapor pressure,
\( i \) is the van't Hoff factor,
\( X \) is the molality of the solute,
\( P^o \) is the vapor pressure of the pure solvent.
For non-electrolytes like urea and cane sugar, the van't Hoff factor (\( i \)) is approximately equal to 1 since these substances do not ionize or dissociate in solution.
For the given solutions:
\( X = \frac{6 \text{ g}}{60 \text{ g/mol} \times 1 \text{ kg}} = 0.1 \text{ mol/kg} \)
\( \Delta P_1 = 1 \times 0.1 \times P^o \)
\( X = \frac{34.2 \text{ g}}{342 \text{ g/mol} \times 1 \text{ kg}} = 0.1 \text{ mol/kg} \)
\( \Delta P_2 = 1 \times 0.1 \times P^o \)
As we can see, both solutions have the same molality, and since the van't Hoff factor is 1 for both substances, the lowering of vapor pressure (\( \Delta P \)) for both solutions will also be the same.
Therefore, the correct answer is same as that of the second solution.