`10.1g" of "KNO_(3)` is dissolved in 500 mL of `H_(2)O`. Mass of `Ba(NO_3)_(2)` that should be added to this solution to get a molality of `0.3` with respect to `NO_(3)^(-)` irons is `(Mw" of "KNO_(3)=101g""mol^(-1), Mw" of "Ba (NO_3)_(2)=261 g""mol^(-1))`

Options

A. `1.38g`
B. 13 g
C. `6.5 g`
D. 65 g

Answer

C

Explanation

Moles of `KNO _(3) = ` moles of `NO _(3) ^(Ө) = (10.1)/(101) = 0.1 `
`because d H _(2) O = 1 g mL ^(-1).` Weight of `H _(2) O = 500 g = 0. 5 kg `
`M _( NO _(3) ^(-)) = ( "Moles of " NO _(3) ^(-))/( "Weight of solvent"(H_(2) O ) "in kg ") = ( n _( NO _(3) ^(-)))/( 0.5 kg)`
`implies 0.3 = ( n _( NO _(3)^(-)))/( 0.5 kg ) , n _( NO _(3) ^(-)) = 0. 3 xx 0.5 = 0. 15`
Moles of `NO _(3) ^(Ө) ` obtained form `KNO _(3) = 0.1 `
Moles of `NO _(3) ^(Ө)` required from `Ba (NO _(3)) _(2) = 0. 15 - 0.1 = 0 .05`
[2 mol of `NO _(3)^(Ө) ` is obtained from 1 mol of `Ba ( N O_(3)) _(2) `]
`therefore `Moles of `Ba ( NO _(3)) _(2) = ( 0. 05)/(2), ` Weight of `Ba ( NO _(3)) _(2) = ( 0.05)/( 2) xx 261 = 6. 5 g`