How many grams of NaOH(80% purity) will be
required to noutralize 12.2 g benzoic acid?
(a) 12.2g
(b) 6g
(c) 5g
(d) 48
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molecular weight of C6H6CO2H= 123gm
so 12.2 gm = appx 0.1 mole
now
1 benzoic acid + 1 NaOH = 1 sodium benzoate + 1 water
so to neutralise 1 mole benzoic acid we need 1 mole NaOH
to neutralise 0.1 mole benzoic acid we need 0.1 mole NaOH
molecular mass of NaOH = 40 gm
so 0.1 mole NaOH =4 gm
purity of sample =80%
100 gm sample contain 80 gm NaOH
to obtain 4 gm NaOH we need 5 gm sample
so 12.2 gm = appx 0.1 mole
now
1 benzoic acid + 1 NaOH = 1 sodium benzoate + 1 water
so to neutralise 1 mole benzoic acid we need 1 mole NaOH
to neutralise 0.1 mole benzoic acid we need 0.1 mole NaOH
molecular mass of NaOH = 40 gm
so 0.1 mole NaOH =4 gm
purity of sample =80%
100 gm sample contain 80 gm NaOH
to obtain 4 gm NaOH we need 5 gm sample
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