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Solution


Q.

X g of benzoic acid on reaction with aq NaHCO3 released CO2 that occupied 11.2 L volume at STP.

X is _______ g.                                        [2025]

Ans.

(61)

Moles of CO2(nCO2)=Volume at STP (in L)22.4=11.222.4=0.5 mol

C6H5COOH+NaHCO3⟶C6H5COONa+H2O+CO2

By stoichiometry of the reaction, moles of benzoic acid (nC6H5COOH)=nCO2=0.5 mol

Molar mass of benzoic acid (MC6H5COOH)
=(12×7+16×2+1×6)  g mol-1=122 g mol-1

Mass of benzoic acid =nC6H5COOH×MC6H5COOH=0.5 mol×122 g mol-1=61 g