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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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Similar questions

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

Q. Two elements 'A' and 'B' form compounds having molecular formulae

A B_{2}

and

A B_{4}

when dissolved in

20.0 g

of benzene,

1.0 g

A B_{2}

lowers the freezing pioint by

{2.3}^{\circ} C

whereas

1.0 g

A B_{4}

lowers the freezing point by

{1.3}^{\circ} C

. The molal depression constant for benzene is

5.1 K k g m o l^{- 1}

. Calculate the atomic masses of A and B.

Q. Two elements form compounds having molecular formula

A B_{2}

and

A B_{4}

. When dissolved in 20 g of benzene, 1.0 g of

A B_{4}

lowers it by 1.3 K. The molar depression constant for benzene is 5.1 K kg

m o l^{- 1}

. Calculate the atomic mass of

A

and

B

Two elements A and B form compounds of formula AB₂ and AB₄. When dissolved in 20.0g of benzene 1.0g of AB₂lowers freezing point by 2.3^°C whereas 1.0g of AB₄ lowers freezing point by 1.3^°C. The K_f for benzene. The atomic masses of A and B are

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Two compounds A and B forms non-volatile compounds having molecular formulae AB2 and AB4. when 1 g of AB2 and 1g of AB4 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. kb of Benzene is S.1 k kg/mol

The correct option is B MA=25.60g,MB=42.65g 2.3=1(MA+2MB)×100020×5.1−(1)1.3=1(MA+4MB)×100020×5.1−(2) Solving (1) & (2) gives MA=25.60 g and MB=42.65g

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 $M_{B} = 42.65 g$