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» 70 Solution :Eq. mass of `CaCO_(3) = (40 + 12 + (3 xx 16))/2 = 50`. Suppose 1.0 g of the sample contains only W g of `CaCO_3`. `THEREFORE` Number of gram EQUIVALENTS of `CaCO_(3)` in 1.0 g sample `=("Mass")/("Eq. mass") = w/50` Number of gram equivalents of HCI reacted = `(N xx V)/1000` `=(1/10 xx 150)/(1000) = 15/1000` According to the law of equivalence, number of gram Eq. of `CaCO_(3) `=no. of gram Eq. of HCl `therefore w/50 = 15/1000` or `w =(50 xx 15)/1000 = 0.75 g` `therefore` Purity of the sample `=0.75/1 xx 100 = 75%`