CBSE Class 11-science Answered
please answer this question:
Asked by prernamehta | 22 Apr, 2010, 10:45: AM
Expert Answer
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 %
Answered by | 22 Apr, 2010, 12:32: PM
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