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In Duma’s method of estimation of nitrogen, 0.35g of an organic compound gave 55ml of nitrogen collected at 300K temperature and 715mm pressure. The percentage composition of nitrogen in the compound would be:
(Aqueous tension at 300K = 15mm)
(A) 16.45%
(B) 17.45%
(C) 14.45%
(D) 15.45%

Answer
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Hint: Using Ideal gas equation, PV=nRT, find a relationship between the given pressure, temperature and volume and at STP. That relation is used in Duma’s method to calculate the volume at STP and then from the molecular weight of nitrogen which is 28g for 22.4L, cross-multiply and find out the weight of nitrogen at STP. Then find out the percentage.

Complete Answer:
Duma’s method is used to quantitatively analyse the percentage of nitrogen in an organic compound. From Duma’s method, we have the formula,
P0V0T0=P1V1T1 where (P1=Pf)
where volume of nitrogen collected =V1 ml
room temperature =T1 K
atmospheric pressure =Pmm
aqueous tension at T1K = f mm
pressure of dry nitrogen =Pf=P1mm
P0, V0 and T0 is the pressure, volume and temperature respectively of dry nitrogen at STP.
From the question,
Given: m=0.35g, P0=760mm, V0=?, T0=273K, P=715mm, f=15mm, V1=55ml, T1=300K
P1=Pf=71515=700mm
P0V0T0=P1V1T1
 V0=P1V1T1×T0P0=700×55300×273760=46.099ml
V0=46.099ml
Now, we know that the molecular weight of nitrogen gas is 28g for 22.4L at STP.
So, 22,400ml of nitrogen at STP weighs 28g.
Therefore, 46.099ml of nitrogen at STP will weigh 28×46.09922,400g
0.35g of organic compound contains 28×46.09922,400g of nitrogen.
Thus, 100g of compound will contain 28×46.099×10022,400×0.35=16.45 of nitrogen.

The percentage of nitrogen in a given organic compound is 16.45%, Option (A).

Note: Remember to convert all units in the same format before doing the calculation. Like in this case, convert all litres to millilitres. Check that all weights are in grams. Remember that molecular weight calculated using atomic mass number is always for 22.4L volume of a gas and for Avogadro’s number of moles.