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When 1 L of 0.1 M sulphuric acid solutio...

When 1 L of `0.1` M sulphuric acid solution is allowed to react with 1 L of `0.1` M sodium hydroxide solution, the amount of sodium sulphate (anhydrous) that can be obtained from the solution fromed and the concentration of `H^(+)` in the solution respectively are :

A

`3.55g,0.1M`

B

`7.10g,0.025M`

C

`3.55g,0.025M`

D

`7.10g,0.05M`

Text Solution

Verified by Experts

The correct Answer is:
D
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Knowledge Check

  • Sulphuric acid reacts with sodium hydroxide as follows : H_(2)SO_(2)+2NaOH - Na_(2)SO_(4)+2H_(2)O When 1 L of 0.1 M sulphuric acid solution is allowed to react will 1L of 0.1 M sodium hydroxide solution, the amount of sodium sulphate formed and its molarity in the solution obtained is

    A
    `"0.1 mol L"^(-1)`
    B
    7.10 g
    C
    `"0.025 mol L"^(-1)`
    D
    3.55 g

  • Sulphuric acid reacts with sodium hydroxide as follows H_(2)SO_(4)+2NaOHrarrNa_(2)SO_(4)+2H_(2)O when 1L of 0.1M sulphuric acid solution is allowed to react with 1L of 0.1M sodium hydroxide solution, the amount of sodium solphate formed and its molarity in the solution obtained is

    A
    `0.1"mol"L^(-1)`
    B
    `7.10g`
    C
    `0.025"mol"L^(-1)`
    D
    3.55g

  • When 1L of 0.1 M sulphuric acid solution is allowed to react with 1 L of 0.1 M sodium hydroxide then the molarity of sodium sulphate formed is (H_(2)SO_(4) + 2NaOH rarr Na_(2)SO_(4) + 2H_(2)O) :

    A
    `0.1 M`
    B
    `0.05 M`
    C
    `0.025 M`
    D
    `0.2 M`

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