

InterviewSolution
Saved Bookmarks
1. |
1 gram of pyrolusite `(MnO_(2))` was boiled with excess of concentrated `HCl` and the issuing gas was passed through a solution of potassium iodide when 1.27 g of iodide were liberated. What is the percentage of pure `MnO_(2)` in the sample ? |
Answer» Calculation of weight of pure `MnO_(2)` The following partial equations are involved in the problem `{:(" "MnO_(2)+4HClrarrMnCl_(2)+Cl_(2)+2H_(2)O),(" "2KI+Cl_(2)rarr2KCl+I_(2)),(bar("Add :"" "underset(55+32(=87g))(MnO_(2))+4HCl+2KIrarrMnCl_(2)+2KCl+2H_(2)underset(2xx127(=254g))(O+I_(2)))):}` 254g of `I_(2)` are obtained from `MnO_(2)=87` g 1.27 g of `I_(2)` are obtained from `MnO_(2)=(87)/(254)xx1.27 g = 0.435 g` Step II. Calculation of percentage purity of the sample Weight of pure `MnO_(2)=0.435 g` Weight of the sample `= 1.0 g` Percentage purity of the sample `= ("Weight of pure MnO"_(2))/("Weight of sample")=((0.435g))/((1.0g))xx100=43.5%`. |
|