JEE Main 12 Jan 2019 Morning Question 30
Question: Freezing point of a 4% aqueous solution of X is equal to freezing point of 12% aqueous solution of Y. If molecular weight of X is A, then molecular weight of Y is [JEE Main Online Paper Held On 12-Jan-2019 Morning]
Options:
A) 2A
B) 3A
C) A
D) 4A
Show Answer
Answer:
Correct Answer: B
Solution:
- $ {{(\Delta T _{f})} _{x}}={{(\Delta T _{f})} _{y}} $ $ K _{f}m _{x}=K _{f}m _{y} $ $ \frac{4\times 1000}{A\times 96}=\frac{12\times 1000}{M _{B}\times 88} $ $ M _{b}=3.27A\simeq 3A $
