Two elements $A$ and $B$ form ideal compounds of formula $A B_{2}$ and $A B_{4}$ . When dissolved in $20.0 g$ of benzene $1.0 g$ of $A B_{2}$ lowers freezing point by ${2.3}^{0} C$ whereas $1.0 g$ of $A B_{4}$ lowers freezing point by ${1.3}^{0} C$ . The $K_{f}$ for benzene is $5.4$ . The atomic masses of $A$ and $B$ are:

27, 45

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42, 25

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52, 48

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48, 52

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Solution

The correct option is A $27, 45$
$Δ T_{f} = - K_{f} m$
For $A B_{2}$ ,
$2.3 = 5.4 \times \frac{1}{(M . ω t)_{1}} \times \frac{1000}{20}$
$(M . ω t)_{1} = \frac{50 \times 5.4}{2.3}$
$(M . ω t)_{1} = 117.39 \Rightarrow M_{A} + 2 M_{B} = 117 - (1)$
Now for $A B_{4}$ ,
$1.3 = 5.4 \times \frac{1}{(M . ω t)_{2}} \times \frac{1000}{20}$
$(M . ω t)_{2} = \frac{5.4 \times 50}{1.3} = 207$
So $M_{A} + 4 M_{B} = 207 - (2)$
Now $(2) - (1)$
$2 M_{B} = 90$

Put in $e q (1)$ ,
$M_{B} = 45$

$M_{A} = 27$

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Similar questions

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

Q. Two elements

A

and

B

form compounds having formula

A B_{2}

and

A B_{4}

. When dissolved in

20 g

of benzene

(C_{6} H_{6}), 1 g

A B_{2}

lowers the freezing point by

2.3 K

whereas

1.0 g

A B_{4}

lowers it by

1.3 K

. The molar depression constant for benzene is

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

. Calculate atomic masses of

A

and

B

QUESTION 2.21

Two elements A and B form compounds having formula $A B_{2}$ and $A B_{4}$ . When dissolved in 20 g of benzene $(C_{6} H_{6})$ , 1 g $A B_{2}$ lowers the freezing point by 2.3 K whereas 1.0 g of $A B_{4}$ lowers it by 1.3K. The molal depression constant for benzene is 5.1 K kg $m o l^{- 1}$ . Calculate atomic masses of A and B.