A) 15.46
B) 16.46
C) 17.46
D) 14.46
Correct Answer: B
Solution :
Given Wt. of compound taken (w)\[=0.35g\] volume of nitrogen collected (V) =55mL room temperature (T)=300Ka Atmospheric pressure (P)=715mm |
Aq. Tension \[\left( \rho \right)\]15=mm |
Calculation ? volume of \[{{\operatorname{N}}_{2}}\]at NTP\[=\frac{\left( P-\rho \right)\times V}{T}\times \frac{273}{760}mL\]\[=\frac{\left( 715-15 \right)\times 55}{300}\times \frac{273}{760}=46.098mL\] |
\[%\,of\,nitrogen\]\[=\frac{28\times vol.of\,{{N}_{2}}at\,NTP\times 100}{22400\times Wt\,of\,organic\,compound}\]\[=\frac{28\times 46.098\times 100}{22400\times 0.35}=16.46%\] |
You need to login to perform this action.
You will be redirected in
3 sec