| What is the mass of precipitate formed when 50 mL of 16.9% solution of \[AgN{{O}_{3}}\] is mixed with 50 mL of 5.8% NaCl solution? [NEET 2015 (Re)] |
| (Ag = 107.8. N =14, O = 16, Na = 23, 01 = 35.5) |
A) 28 g
B) 3.5 g
C) 7 g
D) 14 g
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 |
| \[=0.049\times 143.5=7\,g\] |
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