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Răspuns :

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%