The freezing point of a `3%` (by weight) aqueous solution of `A` is equal to the freezing point of `9%` (by weight) aqueous solution of `B`. If the molecular weight of `A` is `60`, then the molecular weight of `B` will be
a`191` ,b.`90` , c.`45` , d.`20`

Explanation

To solve the problem, we need to use the concept of freezing point depression and molality. The freezing point depression is related to the molality of the solution and the van 't Hoff factor, but since we are not given any information about dissociation, we can assume that both solutes do not dissociate in solution.

### Step-by-step Solution:

1. **Understand the Problem**: We have two solutions: a 3% (by weight) aqueous solution of A and a 9% (by weight) aqueous solution of B. The freezing points of these two solutions are equal. We need to find the molecular weight of B, given that the molecular weight of A is 60.

2. **Freezing Point Depression Formula**: The freezing point depression (ΔTf) can be expressed as:
\[
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