Two elements A and B form compounds having molecular formulae $A B_{2}$ and $A B_{4} .$ When dissolved in $20.0 g$ of Benzene, $1.00 g$ of $A B_{2}$ lowers f.p. by $2 . 3^{\circ} C$ whereas $1.00 g$ of $A B_{4}$ lowers f.p. by $1 . 3^{\circ} C$ . The molal depression constant for benzene is 5.1. The atomic masses of A and B are respectively :

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can't be predicted

42.6, 25.59

30.60

25.59, 42.6

answer is D.

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Detailed Solution

$\begin{array}{l} 2.3 \times (M_{1} + 2 M_{2}) = 1.3 \times (M_{1} + 4 M_{2}) \\ \Rightarrow 0.77 M_{1} = 0.46 M_{2} \\ \Rightarrow 1.67 M_{1} = M_{2} \end{array}$

Now, use, $Δ T_{b} = K_{b} \times M$ .

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