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

Text Solution

Verified by Experts

Applying the formula, `M_(2)=(1000K_(f)w_(2))/(w_(1)xxDeltaT_(f))`
`M_(AB_(2))=(1000xx5.1xx1)/(20xx2.3)"110.87 g mol"^(1),M_(AB_(4))= (1000xx5.1xx1)/(20xx1.3)= "196.15 g mol"^(-1)`
Suppose atomic masses of A and B are 'a' and 'b' respectively. Then
Molar mass of `AB_(2)=a+2b=110.87" g mol"^(-1)" ...(i)"`
Molar mass of `AB_(4)=a+4b="196.15 g mol"^(-1)" ...(ii)"`
Eqn. (ii) - Eqn. (i) gives 2 b = 85.28 or b = 42.64
Substituting in eqn. (i), we get `a+2xx42.64=110.87 or a=25.59`
Thus, Atomic mass of A = 25.59 u, `" "` Atomic mass of B = 42.64 u
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Two elements A and B form compounds having molecular formula AB_(2) and AB_(4) . When dissolved in 20 g of benzene, 1 g of AB_(2) lowers the freezing point by 2.3 K , whereas 1.0 g of AB_(4) lowers it by 1.3 K . The molar depression constant for benzene is 5.1 K kg mol^(-1) . Calculate the atomic mass of A and B .

The elements X and Y form compounds having molecular formula XY_(2) and XY_(4) . When dissolved in 20 gm of benzene, 1 gm XY_(2) lowers the freezing point by 2.3^(@)C , whereas 1 gm of XY_(4) lowers the freezing poing by 1.3^(@) . The molal depression constant for benzene is 5.1. Calculate the atomic masses of X and Y.

Knowledge Check

  • Two elements A and B form compounds having molecular formulae AB_2 and AB_4 when dissolved in 20 g of C_6H_6. 1g AB_2 lowers the freezing point by 2.3 K whereas 1.0 g of AB_4 lowers it by 1.3 K. The molal depression constant for benzene is 5.1 K kg mol^(-1) . The atomic masses of A and B are, respectively

    A
    A = 26, B = 42.64
    B
    A = 31.72, B = 47.02
    C
    A=13.11, B=24.25
    D
    A = 19.17, B = 35.01

  • Two elements A and B form compounds having molecular formula AB_(2) and AB_(4) . When dissolved in 20g of benzene, 1g of AB_(2) lowers the fpt by 2.3K whereas 1g of AB_(4) lowers it by 1.3K (K_(f) for C_(6)H_(6) = 5.1 K m^(-1)) At. Mass of A and B are

    A
    `42.64` and `25.58`
    B
    `25.58` and `42.64`
    C
    `22.64` and `55.58`
    D
    `45.64` and `22.58`

  • The elements X and Y form compound having molecular formula XY_(2) and XY_(4) (both are non - electrolysis), when dissolved in 20 g benzene, 1 g XY_(2) lowers the freezing point by 2.3^(@)c whereas 1 g of XY_(4) lowers the freezing point by 1.3^(C) . Molal depression constant for benzene is 5.1. Thus atomic masses of X and Y respectively are

    A
    42.64, 21.10
    B
    21.10, 42.64
    C
    25.59, 42.64
    D
    42.64, 25.69

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