A) 28g
B) 3.5g
C) 7g
D) 14g
Correct Answer: C
Solution :
Plan For the calculation of mass of \[AgCl\] precipitated, we find mass of \[AgN{{O}_{3}}\] and \[NaCl\]in equal volume with the help of mole concept. 16.9% solution of \[AgN{{O}_{3}}\] means 16.9 g \[AgN{{O}_{3}}\]is present in 100 mL solution. \[\therefore \]8.45 g \[AgN{{O}_{3}}\] will present in 50 mL solution. Similarly, 5.8 g \[NaCl\] is present in 100 mL solution \[\therefore \]2.9 g \[NaCl\] is present in 50 mL solution \[AgN{{O}_{3}}+NaCl\xrightarrow{\,}\,AgCl\,+NaN{{O}_{3}}\] Initial mole \[\frac{8.45}{169.8}\,\] \[\frac{2.9}{58.5}\] 0 0 = 0.049 =0.049 After reaction 0 0 0.049 0.049 \[\therefore \]Mass of \[AgCl\] precipitated \[=0.049\times 143.5=7\,g\]
You need to login to perform this action.
You will be redirected in
3 sec
