A solution is prepared by dissolving 0.63 g of oxalic acid in 100 cm^(3) of water. Find the normality of the solution.

A solution is prepared by dissolving 0.63 g of oxalic acid in 100 cm^(

1.	A solution is prepared by dissolving 0.63 g of oxalic acid in 100 cm^(3) of water. Find the normality of the solution.
Answer» <html><body><p></p>Solution :Molecular mass of oxalic acid, <br/> `(COOH)_(2) . 2H_(2)O = 126` <br/> <a href="https://interviewquestions.tuteehub.com/tag/basicity-394484" style="font-weight:bold;" target="_blank" title="Click to know more about BASICITY">BASICITY</a> of oxalic acid =2 <br/> `therefore` <a href="https://interviewquestions.tuteehub.com/tag/equivalent-446407" style="font-weight:bold;" target="_blank" title="Click to know more about EQUIVALENT">EQUIVALENT</a> mass of oxalic acid `=126/2 = 63` <br/> Number of gram equivalents of oxalic acid <a href="https://interviewquestions.tuteehub.com/tag/dissolved-956358" style="font-weight:bold;" target="_blank" title="Click to know more about DISSOLVED">DISSOLVED</a> `=0.63/63 = 0.01`<br/> Volume of the solution `=100 cm^(3) = 100/1000 L = 0.<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a> L` <br/> <a href="https://interviewquestions.tuteehub.com/tag/normality-1124423" style="font-weight:bold;" target="_blank" title="Click to know more about NORMALITY">NORMALITY</a> of the solution `=("No. of g eq. of oxalic acid")/("Volume in litres")` <br/> `=0.01/0.1 = 0.1 g eq. L^(-1)`</body></html>