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4g of impure CaCO3 on treatment with excess HCl produces 0.88 g CO2. What is the percent purity of CaCO3 sample?

A.75

B.25

C.90

D.50

Answers (1)

best_answer

  CaCO_{3}+2HCl\rightarrow CaCl_{2}+CO_{2}+H_{2}O

here, In 44g of CO_{2} ------> 100g of CaCO_{3}

So, for 0.88g of CO2 -----> (100/44)*0.88 g = 2g 
but in question we have 4g so here we conclude that 2 g is pure CaCO3& 2g is impure.
 

% of purity = 2/4 *100 = 50%

the option C  IS correct

 

Posted by

manish

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