JEE Main 2022 · 24 Jun · Shift 1 · Chemistry Q88

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ChemistryInteger+4 / -1Question 88 of 90

A 0.166 g sample of an organic compound was digested with conc. $\mathrm{H_2SO_4}$ and then distilled with NaOH. The ammonia gas evolved was passed through 50.0 mL of 0.5 N $\mathrm{H_2SO_4}$ . The used acid required 30.0 mL of 0.25 N NaOH for complete neutralization. The mass percentage of nitrogen in the organic compound is ______.

Your answer

Answer and worked solution

Official answer: 63

Reading the data. Taking 7.5 meq of NaOH as the excess acid would leave $25-7.5=17.5$ meq for $\mathrm{NH_3}$ , which is 0.245 g N, more than the 0.166 g sample. So the intended meaning (and the official answer) is that the acid used up by ammonia is equivalent to 30.0 mL of 0.25 N NaOH, i.e. 7.5 meq.

Kjeldahl. $\%\mathrm N=\dfrac{1.4\times\text{meq of acid used}}{w}=\dfrac{1.4\times7.5}{0.166}\approx63.3$ .

Answer: 63

Topics:Qualitative and quantitative analysis of organic compounds Equivalent concept and normality

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JEE Main 2022 · 24 Jun · Shift 1, Chemistry question 88