The molality of an urea solution in which 0.0100 g of urea, [( NH 2)2 CO ] is added to 0.3000 dm 3 of water at STP is

Question

The molality of an urea solution in which 0.0100 g of urea, [( NH 2)2 CO ] is added to 0.3000 dm 3 of water at STP is (A) 0.555 m (B) 5.55 × 10-4 m (C) 33.3 m (D) 3.33 × 10-2 m

Tardigrade Admin · Accepted Answer

Molality =( text moles of solute / kg text of water ) Moles of urea =(0.010/60) mol Water at STP ( d =1 g / cm 3=1 kg / dm 3)=0.3 dm 3=0.3 kg ∴ Molality =(0.010/60 × 0.3)=5.55 × 10-4 m

Q. The molality of an urea solution in which $0.0100 g$ of urea, $[(N H_{2})_{2} CO]$ is added to $0.3000 d m^{3}$ of water at STP is

$0.555 m$

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

$33.3 m$

$3.33 \times 1 0^{- 2} m$

Molality $= \frac{moles of solute}{k g of water}$
Moles of urea $= \frac{0.010}{60} m o l$
Water at STP $(d = 1 g / c m^{3} = 1 k g / d m^{3}) = 0.3 d m^{3} = 0.3 k g$
$∴$ Molality $= \frac{0.010}{60 \times 0.3} = 5.55 \times 1 0^{- 4} m$

Q. The molality of an urea solution in which 0.0100g of urea, [(NH2​)2​CO] is added to 0.3000dm3 of water at STP is

0.555m

5.55×10−4m

33.3m

3.33×10−2m