The amount of $\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}$ (molecular weight 261.32 amu) required to be added to 500 g of a 0.11 mol $\mathrm{kg}^{-1}$ solution of $\mathrm{KNO}_{3}$ in order to raise its ionic strength to 1.00 is approximately
(a)38.8 g
(b)19.4 g
(c)76.2 g
(d)126.5 g
Answer
Answer: A ✓ checked by 4AB · confidence high
Explanation
Need ΔI = 0.89; Ba(NO₃)₂ gives I = 3m → m = 0.297 mol kg⁻¹ × 0.5 kg = 0.148 mol × 261.3 g = 38.8 g (a).