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 :

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 : (A) A (B) 3A (C) 4A (D) 2A

Tardigrade Admin · Accepted Answer

For same freezing point, molality of both solution should be same. mx =my (4×1000/96 × Mx) = (12×1000/88× My) or, My = (96×12/4 ×88) Mx = 3.27 A Closest option is 3A.

Q. 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 :

A

3A

4A

2A

For same freezing point, molality of both solution should be same.
$m_{x} = m_{y}$
$\frac{4 \times 1000}{96 \times M _{x}} = \frac{12 \times 1000}{88 \times M _{y}}$
or, $M_{y} = \frac{96 \times 12}{4 \times 88} M_{x} = 3.27 A$
Closest option is 3A.

Q. 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 :

A

3A

4A

2A

For same freezing point, molality of both solution should be same. mx​=my​ 96×Mx​4×1000​=88×My​12×1000​ or, My​=4×8896×12​Mx​=3.27A Closest option is 3A.

Q. 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 :

For same freezing point, molality of both solution should be same.
$m_{x} = m_{y}$
$\frac{4 \times 1000}{96 \times M _{x}} = \frac{12 \times 1000}{88 \times M _{y}}$
or, $M_{y} = \frac{96 \times 12}{4 \times 88} M_{x} = 3.27 A$
Closest option is 3A.