A) 16.45
B) 17.45
C) 14.45
D) 15.45
Correct Answer: A
Solution :
\[\frac{{{p}_{1}}{{V}_{1}}}{{{T}_{1}}}=\frac{{{p}_{2}}{{V}_{2}}}{{{T}_{2}}}\] Where,\[{{p}_{2}}=\] pressure of \[{{N}_{2}}\] at STP = 760 mm \[{{T}_{2}}\]=Temperature of \[{{N}_{2}}\] at STP = 273 K \[{{V}_{2}}\]? Volume of \[{{\text{N}}_{\text{2}}}\] at STP (By gas equation) \[\left( \frac{\rho -{{\rho }_{1}}}{t+273} \right){{V}_{1}}\times \frac{273}{760}={{V}_{2}}\]where,\[{{\rho }_{1}}=\rho -{{\rho }_{1}}\] \[\rho =715\,\text{mm}\](pressure at which \[{{\text{N}}_{\text{2}}}\] collected). \[{{\rho }_{1}}=\] aqueous tension of water = 15 mm \[{{T}_{1}}=t+273=300K\] \[{{V}_{1}}=55\,mL=\] volume of moist nitrogen in nitro meter \[\therefore \]\[{{V}_{2}}=\frac{(715-15)\times 55}{300}\times \frac{273}{760}=46.098\,mL\] % of nitrogen in given compound \[=\frac{28}{22400}\times \frac{{{V}_{2}}}{W}\times 100=\frac{28}{22400}\times \frac{46.098}{0.35}\times 100\] \[=16.4%\]
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