When 1 g each of compounds AB and AB_2 are d | vedclass

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(D) For $AB$: $\Delta T_b = K_b 	imes m \implies 2.7 = 0.5 	imes \frac{1/M_{AB}}{15 	imes 10^{-3}} \implies M_{AB} = \frac{0.5 	imes 1000}{2.7 	i

Vedclass · Accepted Answer

$25$

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(D) For $AB$: $\Delta T_b = K_b 	imes m \implies 2.7 = 0.5 	imes \frac{1/M_{AB}}{15 	imes 10^{-3}} \implies M_{AB} = \frac{0.5 	imes 1000}{2.7 	imes 15} = \frac{500}{40.5} \approx 12.34 \ g/mol$. For $AB_2$: $\Delta T_b = K_b 	imes m \implies 1.5 = 0.5 	imes \frac{1/M_{AB_2}}{15 	imes 10^{-3}} \implies M_{AB_2} = \frac{0.5 	imes 1000}{1.5 	imes 15} = \frac{500}{22.5} \approx 22.22 \ g/mol$. Let $A$ and $B$ be the atomic masses of elements $A$ and $B$. $A + B = 12.34$ $(I)$ $A + 2B = 22.22$ $(II)$ Subtracting $(I)$ from $(II)$: $B = 22.22 - 12.34 = 9.88$. Substituting $B$ in $(I)$: $A = 12.34 - 9.88 = 2.46$. $A = 24.6 	imes 10^{-1} \approx 25 	imes 10^{-1}$. Thus,the nearest integer is $25$.

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