5 , mol of AB_2 weigh 125 times 10^{-3} , kg | vedclass

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Question

(C) Molar mass is the mass of $1 \, mol$ of a substance.For $AB_2$: Molar mass $= \frac{125 	imes 10^{-3} \, kg}{5 \, mol} = 25 	imes 10^{-3} \, kg

Vedclass · Accepted Answer

$M_A = 5 	imes 10^{-3}$ and $M_B = 10 	imes 10^{-3}$

Vedclass · Answer

(C) Molar mass is the mass of $1 \, mol$ of a substance.For $AB_2$: Molar mass $= \frac{125 	imes 10^{-3} \, kg}{5 \, mol} = 25 	imes 10^{-3} \, kg \, mol^{-1}$.So,$M_A + 2M_B = 25 	imes 10^{-3} \quad (i)$.For $A_2B_2$: Molar mass $= \frac{300 	imes 10^{-3} \, kg}{10 \, mol} = 30 	imes 10^{-3} \, kg \, mol^{-1}$.So,$2M_A + 2M_B = 30 	imes 10^{-3} \quad (ii)$.Subtracting equation $(i)$ from $(ii)$:$(2M_A + 2M_B) - (M_A + 2M_B) = (30 - 25) 	imes 10^{-3}$.$M_A = 5 	imes 10^{-3} \, kg \, mol^{-1}$.Substituting $M_A$ in equation $(i)$:$5 	imes 10^{-3} + 2M_B = 25 	imes 10^{-3}$.$2M_B = 20 	imes 10^{-3}$.$M_B = 10 	imes 10^{-3} \, kg \, mol^{-1}$.

$5 \, mol$ of $AB_2$ weigh $125 \times 10^{-3} \, kg$ and $10 \, mol$ of $A_2B_2$ weigh $300 \times 10^{-3} \, kg$. The molar mass of $A$ $(M_A)$ and molar mass of $B$ $(M_B)$ in $kg \, mol^{-1}$ are:

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