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Two elements A and B form compounds havi...

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`.

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To solve the problem, we need to calculate the atomic masses of elements A and B based on the information provided about their compounds AB₂ and AB₄. We will use the freezing point depression formula and the data given in the question. ### Step-by-Step Solution: 1. **Understand the Freezing Point Depression Formula**: The freezing point depression (\( \Delta T_f \)) is given by the formula: \[ \Delta T_f = K_f \cdot m ...
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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 (C_(6)H_(6)) , 1 g of AB_(2) lowers the freezing point by 2.4 K whereas 1.0 g of AB_(4) lowers it by 1.3 K. The molal depression constant of benzene is "5.1 K kg mol"^(-1) . Calculate atomic masses of A and B.

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.

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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    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.

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