`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. `1.3 g`
B. `13 g`
C. `6.5 g`
D. `65g`

Answer

C

Explanation

Moles of `KNO_(3)` =moles of `NO_(3)^(Theta) =(10.1)/(101)=0.1`
`:." "dH_(2)O=1 g""mL^(-1)" "` Weight of `H_(2)O= 500mL xx1=500 g =0.5 kg`
`m_(NO_(3)^(-)) =(" Moles of "NO_(3))/(" Weight of solvent "(H_(2)O)" in kg ") =(n_NO_(3)^(-))/(0.5kg)`
`implies 0.3=(n_NO_(3)^(-))/(0.5kg)" ":. {:(n_(NO_(3)^(-))),("required"):}=0.3xx0.5=0.15`
Moles of `NO_(3)^(Theta)` obtained from `KNO_(3)=0.1`
Moles of `NO_(3)^(Theta)` required from `Ba(NO_3)_(2)=0.15-0.1=0.05`
[2 mol of `NO_(3)^(Theta)` is obtained from 1 mol of `Ba (NO_3)_(2)`]
Moles of `Ba(NO_3)_(2)=(0.05)/(2)`, Weight of `Ba(NO_3)_(2)=(0.05)/(2)xx261=6.5 g`.