Two elements A and B form compounds with mole | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

$25.59, 42.64$

Vedclass · Answer

(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 $B = 42.64$.

Similar Questions

Ethylene glycol is used as an anti-freezing agent. Calculate the amount of ethylene glycol to be added to $4 \ kg$ of water to prevent it from freezing at $-6 \ ^{\circ}C$. ($K_f$ for water $= 1.85 \ K \ kg \ mol^{-1}$) (molar mass of $(CH_2OH)_2 = 62 \ g \ mol^{-1}$) .......... $g$.

Calculate the molality of a nonvolatile solution if the solution freezes at $-0.95^{\circ}C$ $[K_f \text{ for water} = 1.86 \ K \ kg \ mol^{-1}, \text{ freezing point of water} = 0^{\circ}C]$.

Identify the correct relation between depression in freezing point and the freezing point of a pure solvent.

The solution containing $6 \ g$ urea (molar mass $60$) per $dm^3$ of water and another solution containing $9 \ g$ of solute $A$ per $dm^3$ water freeze at the same temperature. What is the molar mass of $A$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module