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Standard XII

Chemistry

Dalton's Law of Partial Pressures

In Duma's met...

Question

In Duma’s method of estimation of N 0.35 g of an organic compound gave 55mL of nitrogen collected at 300K temperature and 715mm pressure. The % of N in the compound would be (aqueous tension at 300K=15mm)

15.45

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14.45

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17.45

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16.45

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Solution

The correct option is D 16.45
Volume of $N_{2} = \frac{(P - P_{a q t e n}) \times V}{T} \times \frac{273}{760} = \frac{(715 - 15) \times 55}{300} \times \frac{273}{760} = 46.09$
$% o f N = \frac{28}{2400} \times \frac{V \times 100}{w} = \frac{28}{22400} \times \frac{46.09 \times 100}{0.35} = 16.45$

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In Duma’s method of estimation of N 0.35 g of an organic compound gave 55mL of nitrogen collected at 300K temperature and 715mm pressure. The % of N in the compound would be (aqueous tension at 300K=15mm)

The correct option is D 16.45Volume of N2=(P−Paq ten)×VT×273760=(715−15)×55300×273760=46.09 % of N=282400×V×100w=2822400×46.09×1000.35=16.45

The correct option is D 16.45
Volume of $N_{2} = \frac{(P - P_{a q t e n}) \times V}{T} \times \frac{273}{760} = \frac{(715 - 15) \times 55}{300} \times \frac{273}{760} = 46.09$
$% o f N = \frac{28}{2400} \times \frac{V \times 100}{w} = \frac{28}{22400} \times \frac{46.09 \times 100}{0.35} = 16.45$