Two compounds A and B forms non-volatile compounds having molecular formulae AB 2 and AB 4- when 1 g of AB 2 and 1 g of AB 4 are separately dissolved in 20 g of Benzene- the B-P- was lowered by 2-3- C and 1-3- C respectively- Find the molar mass of A and B- k b of Benzene is S -1 k kg - molA- M A -32 g - B B -26-4 gB- M A -14-6 g - M B -22-4 gC- M A -8-4 g - M B -15 gD- M A -25-60 g - M B -42-65 g

The correct option is D M_A =25.60 g, M_B =42.65 g 2.3 =1/(M_A+2M_B)×1000/20× 5.1 (1) 1.3 =1/(M_A +4M_B)×1000/20× 5.1 (2) Solving (1) amp; (2) gives M_A ...

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Standard XII

Chemistry

Elevation in Boiling Point

Two compounds...

Question

Two compounds A and B forms non-volatile compounds having molecular formulae $A B_{2}$ and $A B_{4}$ . when 1 g of $A B_{2}$ and 1g of $A B_{4}$ are separately dissolved in 20 g of Benzene, the B.P. was lowered by ${2.3}^{\circ} C$ and ${1.3}^{\circ} C$ respectively. Find the molar mass of A and B. $k_{b}$ of Benzene is S.1 k kg/mol

$M_{A} = 32 g, B_{B} = 26.4 g$

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$M_{A} = 25.60 g, M_{B} = 42.65 g$

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$M_{A} = 14.6 g, M_{B} = 22.4 g$

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$M_{A} = 8.4 g, M_{B} = 15 g$

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Solution

The correct option is B
$M_{A} = 25.60 g, M_{B} = 42.65 g$

$2.3 = \frac{1}{(M_{A} + 2 M_{B})} \times \frac{1000}{20} \times 5.1 - (1) 1.3 = \frac{1}{(M_{A} + 4 M_{B})} \times \frac{1000}{20} \times 5.1 - (2)$
Solving (1) & (2) gives
$M_{A} = 25.60$ g and $M_{B} = 42.65 g$

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