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Calculate the molality of 1 litre soluti...

Calculate the molality of 1 litre solution of 93% `H_(2)SO_(4) ("weight"//"volume")`. The density of solution is 1.84 g `mL^(-1)`.

Text Solution

Verified by Experts

The solution is `1000mL` containing `930g` of `H_(2)SO_(4)`
The weight of the solution will be `1840g`. The weight of the solvent `(H_(2)O)` will therefore be `(1840-930)` i.e.910g
`:.` molality `=("mole of" H_(2)SO_(4))/("wt.of"H_(2)O(g))xx1000`
`=(930//98)/(910)xx1000=10.428m`
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Calculate the molality of 1 litre solution of 93% H_(2)SO_(4) (weight/volume). The density of the solution is "1.84 g mL"^(-1)

Calculate the molality of 1 litre solution of 93% H_(2)SO_(4) (weight / volume). The density of the solution is "1.84 g mL"^(-1) .

Knowledge Check

  • If the density of a 1-ltre solution of 98% H_(2)SO_(4) (wt.//vol.) is 1.88 g mol^(-1) , the molality of the solution will be

    A
    `13.13`
    B
    `10.10`
    C
    `11.11
    D
    `12.12`

  • If 1 litre aq. Solution of H_(2)SO_(4) ( 10% by weight) with density 1.1 gm/ml, find th pH of solution.

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    `-log2.244`
    B
    `-log1.96`
    C
    `-log1.86`
    D
    `-log3.87`

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