Question

Determine the amount of urea (NH₂CONH₂) to be added in 1000 g of water to decrease its vapour pressure by 25%. 

Answer 1

Correct Answer: 1111

Given :

Mole concept is used here, and the relationship is: 
\( P_o - P_s = \frac{n}{N+n} \)

• \( P_o \): Initial vapor pressure
• \( P_s \): Vapor pressure of the solution
• \( N \): Number of moles of solvent
• \( n \): Number of moles of solute

Step-by-Step Calculation:

1. The equation simplifies to: \[ 4n = N + n \]
2. Rearranging, we get: \[ n = \frac{N}{3} \]
3. \( N \), the number of moles of solvent (water), is calculated as: \[ N = \frac{1000}{18} \] \[ N = 55.56 \]
4. Now, substituting \( N \) into the formula for \( n \): \[ n = \frac{55.56}{3} = 18.52 \]
5. The weight of urea is calculated as: \[ \text{Weight of urea} = n \times \text{Molar mass of urea (60 g/mol)} \] \[ \text{Weight of urea} = 18.52 \times 60 = 1111.1 \, \text{g} \]

Conclusion:

The amount of urea is approximately 1111.1 g.

Answer 2

\(\frac{P^0-P_s}{P_s}=\frac{n_{solute}}{n_{solvent}}\)
= \(\frac{\frac{x}{60}}{\frac{1000}{18}}=\frac{P^0-0.75P^0}{0.75P^0}\)
\(⇒x=\frac{10000}{9}=1111\ gm\)
So , the correct answer is 1111 gm