ms.NaOH = 20 g , c% = 10%
c% = mdx100/ms => md = msxc%/100
= 20x10/100 = 2 g NaOH
m g 2 g
C6H5COOH + NaOH --> C6H5COONa + H2O
122 40
=> m = 122x2/40 = 6,1 g acid benzoic pur
100% ....................... 10g acid impur
p% ............................ 6,1 g acid pur
= 61%
