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A 20.00 ml sample of Ba (OH)2 solution i...

A 20.00 ml sample of `Ba (OH)_2` solution is titrated with 0.245 M HCI. If 27.15 ml of HCI is required, then the molarity of the `Ba (OH)_2` solution will be :

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Milli eq. HCl initially ` = 10 xx 0.5 = 5 `
Milli eq. of NaOH consumed = Milli eq. of HCl in excess = `10 xx 0.2= 2`
` :. `Milli eq. of HCl consumed = Milli eq. of `Ba(OH)_(2) = 5 - 2 = 3`
`:. " Eq. of " Ba(OH)_(2) = 3//1000 = 3 xx 10^(-3)`
Mass of `Ba(OH)_(2) = 3 xx 10^(-3 xx) (171 //2) = 0.2565 `g
` % Ba(OH)_(2) = (0.2565//10) xx 100 = 2.56 %`
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