99 Percentile Qs Bank for JEE MainChemistrySome Basic Concepts of Chemistry
A piece of plumber's solder (containing P b & S n only) weighing 3   g m was dissolved in dil H N O 3 and then treated with dilute H 2 S O 4 - the p p t . of P b S O 4 is produced , which after washing and drying weighed 2.93   g m . The remaining solution was neutralized - a white ppt of stannic acid is produced , which on heating yields 1.27   g m of S n O 2 . Therefore mass % of P b and Sn in th
Options
- A% P b = 66.67 % , % S n = 33.33 %
- B% P b = 33.33 % , S n = 66.67 %
- C% P b = 50 % , % S n = 50 %
- D% P b = 75 % , % S n = 25 %
Correct answer
A. % P b = 66.67 % , % S n = 33.33 %
Step-by-step solution
wt   of   Pb = Wt .   of   Pb Wt .   of   PbSO 4 × 2 . 93 w t   o f   P b = 207 303   ×   2.93 = 2   g m if sample weight is 3 gm than wt of S n = 1   g m . ∴ %   P b = 2 3   ×   100 = 66.67   % ;   h e n c e   %   S n = 100   –   66.67 = 33.33   %