Two elements A and B form compounds with mole | vedclass

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(B) Let $a$ and $b$ be the atomic weights of $A$ and $B$ respectively.For $AB_2$,molar mass $M_1 = a + 2b$.For $AB_4$,molar mass $M_2 = a + 4b$.Using

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$25.59, 42.64$

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(B) Let $a$ and $b$ be the atomic weights of $A$ and $B$ respectively.For $AB_2$,molar mass $M_1 = a + 2b$.For $AB_4$,molar mass $M_2 = a + 4b$.Using the formula $\Delta T_f = \frac{1000 	imes K_f 	imes w}{M 	imes W}$:For $AB_2$: $2.3 = \frac{1000 	imes 5.1 	imes 1}{(a + 2b) 	imes 20} \implies a + 2b = \frac{5100}{2.3 	imes 20} \approx 110.87 \dots (1)$For $AB_4$: $1.3 = \frac{1000 	imes 5.1 	imes 1}{(a + 4b) 	imes 20} \implies a + 4b = \frac{5100}{1.3 	imes 20} \approx 196.15 \dots (2)$Subtracting $(1)$ from $(2)$: $(a + 4b) - (a + 2b) = 196.15 - 110.87 \implies 2b = 85.28 \implies b = 42.64$.Substituting $b$ in $(1)$: $a + 2(42.64) = 110.87 \implies a + 85.28 = 110.87 \implies a = 25.59$.Thus,the atomic weights are $A = 25.59$ and $B = 42.64$.

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