How many grams (to the nearest 0.01 g) of NH4Cl (Mm = 53.49 g/mol) must be added to 800. mL of 1.202-M solution?

NH₃(aq) + H₂O(l) ⇌ NH₄⁺(aq) + OH⁻(aq) …. Kb = 1.8 × 10⁻⁵

pOH = pKb - log([NH₃]/[ NH₄⁺])
log([NH₃]/[ NH₄⁺]) = pKb - pOH
[NH₃]/[ NH₄⁺] = 10^(pKb - pOH)
[ NH₄⁺] = [NH₃] / 10^(pKb - pOH)
[ NH₄⁺] = 1.202 / 10^{-log(1.8 × 10⁻⁵) - (14 - 9.30)}
[ NH₄⁺] = 1.084 M

No. of moles of NH₄⁺ ions = (1.084 mol/L) × (800/1000 L) = 0.8672 mol
No. of moles of NH₄Cl added = 0.8672 mol
Mass of NH₄Cl added = (0.8672 mol) × (53.49 g/mol) = 46.39 g