A) 16.76
B) 15.76
C) 17.36
D) 18.20
Correct Answer: A
Solution :
Mass of organic compound = 0.25 g Experimental values At STP \[{{V}_{1}}=40\,mL\] \[{{V}_{2}}=?\] \[{{T}_{1}}=300K\] \[{{T}_{2}}=273\,K\] \[{{P}_{1}}=725-25=700\,mm\] \[{{P}_{2}}=760\,mm\] \[\frac{{{P}_{1}}{{V}_{1}}}{{{T}_{1}}}=\frac{{{P}_{2}}{{V}_{2}}}{{{T}_{2}}}\] \[{{V}_{2}}=\frac{{{P}_{1}}{{V}_{1}}{{T}_{2}}}{{{T}_{1}}{{P}_{2}}}=\frac{700\times 40\times 273}{300\times 760}=33.52\,mL\] 22400 mL of \[{{N}_{2}}\]at STP weights = 28 g \[\therefore \]33.52 mL of \[{{N}_{2}}\] at STP weighs \[=\frac{28\times 33.52}{22400}=0.0419g\] % of \[N=\frac{\text{Mass}\,\text{of}\,\text{nitrogen}\,\text{at}\,\text{STP}}{\text{Mass}\,\text{of}\,\text{organic}\,\text{compound}\,\text{taken}}\times 100\] \[=\frac{0.0419}{0.25}\times 100=16.76%.\]
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