Question

What weight of the non-volatile solute, urea (NH₂ – CO – NH₂) needs to be dissolved in 100g of water, in order to decrease the vapour pressure of water by 25%? What will be the molality of the solution?

Answer 1

Tell me if I am right or not

Answer 2

Hey dear,

● Answer -
Weight of urea required => 83.33 g

● Explanation -
# Given -
W1 = 100 g
W2 = ?
M1 = 18 g
M2 = 60 g
∆P/P° = 25%

# Solution -
Lowering of vapor pressure by non-volatile solute is formulated by -
∆P/P° = W2M1 / W1M2
25/100 = (W2×18) / (100×60)
W2 = 83.33 g

Therefore, 83.33 g of urea is required to lower the vapour pressure.


Hope this helps you..