UncategorizedIf 0.5 g of a mixture of two metals A and B with respective equivalent weights 12 and 9 displace 560 mL of H2 at STP from an acid, the composition of the mixture is

If 0.5 g of a mixture of two metals A and B with respective equivalent weights 12 and 9 displace 560 mL of H2 at STP from an acid, the composition of the mixture is

  1. A

    40% of A and 60% of B

  2. B

    60% of A and 40% of B

  3. C

    30% of A and 70% of B

  4. D

    70% of A and 30% of B

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    Solution:

    According to STP conditions, one mole of a gas has a volume of 22400 mL. At STP conditions, one mole of H2 has a volume of 22400 mL. 

    If one mole of H2 (or 2 equivalents of H2) has a volume of 22400 mL, then one equivalent of H2 has a volume of 11200 mL.

    If one equivalent of H2 has a volume of 11200 mL, then the number of equivalents of H2 present in 560 mL is calculated as:

    11200 mL of H2=One equivalent of H2560 mL of H2=560 mL of H2×One equivalent of H211200 mL of H2 =120 equivalents of H2

    Total mass of the mixture is 5.6 g. Assume that the mass of A is x g and the mass of B is 0.5-x g.

    Equivalent mass of A is calculated as:

    Number of equivalents of A=Given massEquivalent mass of A =x12

    Equivalent mass of B is calculated as:

    Number of equivalents of B=Given massEquivalent mass of B =0.5-x9

    Equivalents of A and B displaces the equivalents of H2. This implies that sum of equivalents of A and B is equal to the equivalents of H2 present in 560 mL.

    Substitute the values in the above equation and determine the value of x.

    x12+0.5-x9=1203x+4(0.5-x)36=120x=0.2

    Therefore, mass of A is 0.2 g and the mass of B is 0.3 g (0.5 g-0.2 g).

    Composition of A in the mixture is calculated as:

    Composition of A in mixture=Mass of ATotal mass of the mixture×100 =0.20.5×100 =40%

    Composition of B in the mixture is calculated as:

    Composition of B in mixture=Mass of BTotal mass of the mixture×100 =0.30.5×100 =60%

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