The amount of Ba(NO₃)₂ (molecular weight 261.32 amu) required to be added to 500 g of a 0.11 mol kg^-1 solution of KNO₃ in order to raise its ionic strength to 1.00 is approximately:

Concept: 

{Ionic strength (I) = \(\frac{1}{2} \sum_{i = no. \ of \ ions} c_i z_i^2\)}

where, c_i = concentration of ions

z_i = charge of ions

Explanation:

In the given question:

\(I = I_{KNO_3} + I_{Ba(NO_3)_2}\)     ....(1)

→ for \(I_{KNO_3}\)_,

\(KNO_3 \ \ \ \ \ \ → K^\oplus + NO_3^\ominus \\\ 0.11 moles \ \ \ \ \ 0.11 \ \ \ \ \ 0.11\)

_∴ \(I = \frac{1}{2} \sum c_i z_i^2\)

I = 1/2 [0.11 × (+1)² + 0.11 × (-1)²]

\(I_{KNO_3} = 0.11\)

Now for \(I_{Ba(NO_3)_2}\)

Here is the question concentration is given in terms of molality. Therefore Let the molality of Ba(NO₃)₂ be "m".

∴ \(Ba(NO_3)_2 \ → Ba^{2+} + 2 NO_3^- \\\ m \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ m \ \ \ \ \ \ \ \ \ \ \ 2m\)

∴ \(I_{Ba(NO_3)_2} = \frac{1}{2} [m(2)^2 + 2m(-1)^2] \)

\(I_{Ba(NO_3)_2} = \frac{1}{2} [4m + 2m]\)

\(I_{Ba(NO_3)_2} = \frac{1}{2} \times 6 m = 3 m \)    ......(2)

Again,

\(\rm molality (m) = \frac{no. \ of \ moles \ of \ solute}{Vol \ of \ solvent \ (kg)}\)

\(\rm ( \because no. \ of \ moles \ of \ solute \rm = \frac{mass \ of \ solute}{molecular \ mass}\)

let mass of solute be 'x'

Given molecular mass = 261.32 amu

∴ m(Ba(NO₃)₂) = \(\frac{x}{261.32} \times \frac{1000}{500}\)

m(Ba(NO3)2) = \(\frac{x}{130.66} \)   ....(2)

From equation (2) and (3),

\(I_{Ba(NO_3)_2} = \frac{3x}{130.66}\)    ....(4)

From equation (1) and (4) we have,

\(I = \frac{3x}{130.66} + 0.11\)    ....(5)

Now according to question the value of 'x' or the amount of Ba(NO₃)₂ added should raise the Ionic strength (I) to 1,

∴ From equation (5),

→ \(1 = \frac{3x}{130.66} + 0.11\)

→ \(\frac{3x}{130.66} = 1-0.11\)

→ \(x = \frac{130.66 \times 0.89}{3}\)

x = 38.76 ≈ 38.8 g

∴ option '1' is correct.

Concsluaion:-

The amount of Ba(NO₃)₂ required to be added is 38.8 g.