150 mL of N/10HCI are required to react completely with 1.0 g of a sample of lime stone (CaCO_3). Calculate the percentage purity of the sample.

150 mL of N/10HCI are required to react completely with 1.0 g of a sam

1.	150 mL of N/10HCI are required to react completely with 1.0 g of a sample of lime stone (CaCO_3). Calculate the percentage purity of the sample.
Answer» <html><body><p>70</p>Solution :Eq. mass of `CaCO_(3) = (40 + <a href="https://interviewquestions.tuteehub.com/tag/12-269062" style="font-weight:bold;" target="_blank" title="Click to know more about 12">12</a> + (3 xx 16))/2 = 50`. <br/> Suppose 1.0 g of the sample contains only <a href="https://interviewquestions.tuteehub.com/tag/w-729065" style="font-weight:bold;" target="_blank" title="Click to know more about W">W</a> g of `CaCO_3`.<br/> `<a href="https://interviewquestions.tuteehub.com/tag/therefore-706901" style="font-weight:bold;" target="_blank" title="Click to know more about THEREFORE">THEREFORE</a>` Number of gram <a href="https://interviewquestions.tuteehub.com/tag/equivalents-974752" style="font-weight:bold;" target="_blank" title="Click to know more about EQUIVALENTS">EQUIVALENTS</a> of `CaCO_(3)` in 1.0 g sample `=("Mass")/("Eq. mass") = w/50` <br/> Number of gram equivalents of HCI reacted = `(N xx V)/1000` <br/> `=(1/10 xx 150)/(1000) = 15/1000` <br/> According to the law of equivalence, number of gram Eq. of `CaCO_(3) `=no. of gram Eq. of HCl <br/> `therefore w/50 = 15/1000` or `w =(50 xx 15)/1000 = 0.75 g` <br/> `therefore` Purity of the sample `=0.75/1 xx 100 = 75%`</body></html>