The molality of a urea solution in which 0.0200 g of urea (NH2CONH2) is added to 0.400 dm3 of water at STP is

Question

The molality of a urea solution in which 0.0200 g of urea (NH2CONH2) is added to 0.400 dm3 of water at STP is (A) 0.555 molal (B) 5.55 × 10-4 molal (C) 8.33 × 10-4 molal (D) 33.3 molal

Tardigrade Admin · Accepted Answer

1 dm3=1000 cm3=1L=1kg 0.400 dm3 of H2O at stp =0.400 Kg of H2O (density = 1.0 g cm ) Moles of urea =(0.0200/60) ∴ Molality =( textMoles of solute (urea) / text kg of solvent) =(0.020/60/0.400) =8.33 × 10-4 mol kg-1

Q. The molality of a urea solution in which 0.0200 g of urea $(N H_{2} CON H_{2})$ is added to $0.400 d m^{3}$ of water at STP is

$0.555$ molal

$5.55 \times 1 0^{- 4}$ molal

$8.33 \times 1 0^{- 4}$ molal

$33.3$ molal

$1 d m^{3} = 1000 c m^{3} = 1 L = 1 k g$
$0.400 d m^{3}$ of $H_{2} O$ at stp
$= 0.400 K g$ of $H_{2} O$
(density = 1.0 g cm )
Moles of urea $= \frac{0.0200}{60}$
$∴$ Molality $= \frac{Moles of solute (urea)}{kg of solvent}$
$= \frac{0.020/60}{0.400}$
$= 8.33 \times 1 0^{- 4} m o l k g^{- 1}$

Q. The molality of a urea solution in which 0.0200 g of urea (NH2​CONH2​) is added to 0.400dm3 of water at STP is

0.555 molal

5.55×10−4 molal

8.33×10−4 molal

33.3 molal