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 the same temperature. The lowering of vapour pressure in the first solution is

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

Same as that of 2nd solution

Double that of 2nd solution

Nearly five times that of 2nd solution

Nearly one-fifth of the 2nd solution

answer is A.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

We know that in the first solution number of the moles of $urea = \frac{Mass of urea}{m . wt . of urea} \times \frac{1}{V} = \frac{12}{60} \times \frac{1}{1} = 0 .2 and,$

In second solution the number of moles of cane sugar $= \frac{Mass of cane sugar}{m . wt . of cane sugar} = \frac{68 .4}{342} \times \frac{1}{1} = 0 .2$

Watch 3-min video & get full concept clarity

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET

How to Join IIT after 10th

🔥 Start Your JEE/NEET Prep at Just ₹1999 / month - Limited Offer! Check Now!

Get Expert Academic Guidance – Connect with a Counselor Today!

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET

How to Join IIT after 10th