The correct option is C 13.2 g
Given, pH=9, Kb(NH3)=2×10−5, [NH3]=0.02 M, V=500 ml
let 'a' millimoles of NH+4 is added to a solution having millimoles
NH4OH=500×0.2=100∴[NH+4]=[salt]=a500 and [NH4OH]=[Base]=100500
pH+pOH=14⇒pOH=14−9=5
Using Henderson equation,
pOH=−logKb+log[NH+4][NH4OH] 5=−log2×10−5+loga500100500
5=5−log2+loga500100500
log2=loga500100500
∴a=200 millimoles = 0.2 mol
(NH4)2SO4 added=a2=0.1 mol ∴W(NH4)2SO4=number of mole×molar mass of (NH4)2SO4
=0.1×132=13.2 g