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

Question

The molality of a urea solution in which 0.0100 g of urea, [(NH2)2CO] is added to 0.3000 dm3 of water at STP is (A) 5.55 × 10-4 M (B) 33.3 M (C) 3.33 × 10-2 M (D) 0.555 M

Tardigrade Admin · Accepted Answer

Molality = ( textmoles of the solute/ textmass of the solvent in kg) Moles of urea (nurea) = (0.0100 g/6. g mol-1) Mass of 1000 mL of solution = volume × density = 300 mL × (1 g/mL) [ ∵ 1 dm3 = 1000 mL] = 300 g Mass of solvent = 300 g - 0.0100 g = 299.9 g = 0.2999 kg Molality = (0.0100/60 times 0.2999) = 5.55 × 10-4 mol kg-1

Q. The molality of a 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

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

$33.3 M$

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

$0.555 M$

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

5.55×10−4M

33.3M

3.33×10−2M

0.555M

Molality =mass of the solvent in kgmoles of the solute​ Moles of urea (nurea​)=6.gmol−10.0100g​ Mass of 1000mL of solution = volume × density =300mL×mL1g​ [∵1dm3=1000mL] =300g Mass of solvent =300g−0.0100g =299.9g=0.2999kg Molality =60 times0.29990.0100​ =5.55×10−4molkg−1

Q. The molality of a 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

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

$33.3 M$

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

$0.555 M$