Pkb = -logKb
       =
-log1.8*10-5
        = 4.75
PHÂ Â = 8.66
POH = 14-PH
          =
14-8.66Â Â = 5.34
no of moles of NH3 = molarity * volume in L
                                Â
= 0.8*2.4 = 1.92 moles of NH3
POHÂ Â Â = Pkb + log[NH4Cl]/[NH3]
5.34Â Â = 4.75 + log[NH4Cl]/1.92
log[NH4Cl]/1.92Â Â Â = 5.34-4.75
log[NH4Cl]/1.92Â Â = 0.59
[NH4Cl]/1.92Â Â Â Â = 100.59
[NH4Cl]/1.92Â Â Â Â = 3.8904
[NH4Cl]Â Â Â Â Â Â Â Â Â Â Â
= 3.8904*1.92Â Â = 7.4695 moles
mass of NH4Cl  = no of moles * gram molar mass
                         Â
= 7.4695*53.5Â Â = 399.62g >>>> answer