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100mL of HC1 gas at 25^(@)C and 740mm pr...

`100mL` of `HC1` gas at `25^(@)C` and `740mm` pressure is dissolved in `1L` of `H_(2)O`. Calculate the `pH` of solution. Given vapour presure of `H_(2)O` at `25^(@)C` is `23.7mm`.

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The correct Answer is:
A, B, D
For `HC1 gas PV = (W)/(Mw) RT`
`:. P_(HC1_(dry)) = (740-23.7)/(760) atm, V = (100)/(1000)L`
`:. T = 25 + 273 = 298 K`
`:.` Mole of `Hc1`, i.e.,
`n_(Hc1) = (W)/(Ww) = (PV)/(RT) = ((740-23.7)xx100)/(760xx1000xx0.0821xx298)`
`:. N = 3.85 xx 10^(-3)`
Now, molarity of `HC1 = (n)/(V_(solution)(L)) = (3.85xx10^(-3))/(1)`
`:. [H^(+)] = 3.85 xx 10^(-3)M`
`:. pH =- log [H^(o+)] =- log 3.85 xx 10^(-3)`
`pH = 2.4142`
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Knowledge Check

  • 100 mL of HCl gas at 25^@ C and 740 mm pressure was dissolved in 1 L of water. What will be the pH of solution. (Given, vapour pressure of H_2 O at 25^@ C is 23.7 mm.)

    A
    `2.012`
    B
    `3.241`
    C
    `2.414`
    D
    `3.021`

  • A sample of H_(2) is collected over H_(2)O at 23^(@)C and a pressure of 732 mm Hg has a volume of 245mL what volume would the dry H_(2) occupy of 0^(@)C and 1 atm pressure? [VP of H_(2)O at 23^(@)C =21 mm Hg]

    A
    211mL
    B
    218mL
    C
    224mL
    D
    249mL

  • A 225 mL sample of H_(2) is collected over water at 25^(@)C and 735mm Hg pressure. Which expression represents the set-up to find the volume of dry H_(2) at 0^(@)C and 1 atmospehre? [VP of H_(2)O at 25^(@)C =24mm of Hg)

    A
    `V=(225xx(735-24)xx273)/(760xx298)`
    B
    `V=(225xx760xx298)/((735-24)xx273)`
    C
    `V=(225xx273xx760)/((735+24)xx298)`
    D
    `V=(225xx(735+24)xx298)/(760xx273)`
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