SATHEE: Solutions Question 246

Solutions Question 246

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

Options:

A) $0.555, m o l, k g^{- 1}$

B) $5.55 \times 10^{- 4}, m o l, k g^{- 1}$

C) $8.33 \times 10^{- 4}, m o l, k g^{- 1}$

D) $33.3, m o l, k g^{- 1}$

Show Answer

Answer:

Correct Answer: C

Solution:

$1, d m^{3} = 100, c m^{3} = 1, L = 1, k g$

$0.400, d m^{^{3}}$ of $H_{2} O$ at STP= $0.400, k g$ of $H_{2} O$ (Density=1.0 g $c m^{- 3}$ ) Moles of urea = $\frac{0.0200}{60} m o l$ (Molar mass of $N H_{2}, C O N H_{2} = 60 g m o l^{- 1}$ )

$∴$ Molality $= \frac{M o l e s, o f, s o l u t, (urea)}{k g, o f, s o l v e n t} = \frac{0.020 / 60}{0.400 k g}$

$= 8.33 \times 10^{- 4} m o l, k g^{- 1}$

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Solutions Question 246

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

Options:

Answer:

Solution:

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