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0.5 g of Ba(OH)2, 0.01 mol Ba(OH)(2) an...

0.5 g of `Ba(OH)_2, 0.01 mol Ba(OH)_(2) ` and 0.01 eq. `Ba(OH)_(2)` were together diluted to one litre. Calculate the normality of basic solution.

Text Solution

Verified by Experts

The correct Answer is:
`0.036 eq L^(-1)`
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Knowledge Check

  • The P^(H) of 0.005 M Ba (OH)_(2) is

    A
    2.301
    B
    11.699
    C
    12
    D
    7

  • 1 mole Ba(OH)_(2) will exactly neutralize

    A
    0.5 mole HCl
    B
    1 mole of `H_(2)SO_(4)`
    C
    `1 mole of `H_(3)PO_(3)`
    D
    3 mole `H_(3)PO_(2)`

  • A solution of 0.1 mole of CH_3 NH_2 ( K_b =5 xx 10^(-4)) and 0.08 mole of HCl is diluted to one litre, then the pOH of the solution is (log 1.25 =0.1)

    A
    10.1
    B
    3.9
    C
    4.9
    D
    9.9

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