SATHEE: Organic Chemistry Some Basic Principles and Techniques

Organic Chemistry Some Basic Principles and Techniques - Result Question 52

####58. In Duma’s method of estimation of nitrogen $0.35 g$ of an organic compound gave $55 m L$ of nitrogen collected at $300 K$ temperature and $715 m m$ pressure. The percentage composition of nitrogen in the compound would be : (Aqueous tension at $300 K = 15 m m$ )

[2011]

(a) 15.45

(b) 16.45

(d) 14.45

Show Answer

Solution:

(b) Given wt of compound taken $(w) = 0.35 g$

Volume of nitrogen collected $(V) = 55 m L$

Room temperature $(t K) = 300 K$

Atmospheric pressure $(P) = 715 m m$

Aq. tension $(ρ) = 15 m m$

Calculation -

Volume of $N_{2}$ at NTP $= \frac{(P - ρ) \times V}{t} \times \frac{273}{760} m L$

$= \frac{(715 - 15) \times 55}{300} \times \frac{273}{760} = 46.098 m L$

of nitrogen

$= \frac{28 \times vol. of N_{2} at N T P \times 100}{22400 \times wt of organic compound}$

$= \frac{28 \times 46.098 \times 100}{22400 \times 0.35} = 16.46$

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Organic Chemistry Some Basic Principles and Techniques - Result Question 52

Solution:

Calculation -

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