please answer this question:

        NaCl   +    AgNO3         AgCl    +   NaNO3

MM = 58.5 g                                  MM = 143.5 g

1 mole of NaCl gives 1 mole of AgCl

58.5 g of NaCl gives 143.5 g of AgCl

0.50 g of NaCl gives = (143.5g * 0.50 g) / 58.5 g of AgCl

                                     = 1.23 g of AgCl

But the amount of AgCl precipitated was 0.90 g.

Percentage purity = ( 0.90 g / 1.23 g) * 100 = 73.2 %