SATHEE: Solutions Question 60

Solutions Question 60

Question: A solution is obtained by dissolving 12 g of urea (mol.wt.60) in a litre of water. Another solution is obtained by dissolving 68.4 g of cane sugar (mol.wt. 342) in a litre of water at are the same temperature. The lowering of vapour pressure in the first solution is [CPMT 2001]

Options:

A) Same as that of 2nd solution

B) Nearly one-fifth of the 2nd solution

C) Double that of 2nd solution

D) Nearly five times that of 2nd solution

Show Answer

Answer:

Correct Answer: A

Solution:

We know that in the first solution number of the moles of urea $= \frac{Mass of urea}{m . w t . of urea} \times \frac{1}{V}$

$= \frac{12}{60} \times \frac{1}{1} = 0.2$ and In second solution the number of moles of cane sugar $= \frac{Mass of cane sugar}{m . w t . of cane sugar}$

$= \frac{68.4}{342} \times \frac{1}{1} = 0.2$ .

Solutions Question 60

Question: A solution is obtained by dissolving 12 g of urea (mol.wt.60) in a litre of water. Another solution is obtained by dissolving 68.4 g of cane sugar (mol.wt. 342) in a litre of water at are the same temperature. The lowering of vapour pressure in the first solution is [CPMT 2001]

Options:

Answer:

Solution:

Engineering Preparation

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NCERT Books & Solutions

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NCERT Exercise Video Solution

Solutions Question 60

Question: A solution is obtained by dissolving 12 g of urea (mol.wt.60) in a litre of water. Another solution is obtained by dissolving 68.4 g of cane sugar (mol.wt. 342) in a litre of water at are the same temperature. The lowering of vapour pressure in the first solution is [CPMT 2001]

Options:

Answer:

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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