JEE Main · 2021 · 25 Jul 2021 · Shift 2 · Practical and Analytical Chemistry

The nitrogen in the sample weighs $\frac{42}{100}\times 0.8=0.336$ g, which is $\frac{0.336}{14}=0.024$ mol. In Kjeldahl's method every nitrogen atom of the compound is converted to one molecule of ammonia, so 0.024 mol of $\mathrm{N}{\mathrm{H}}_3$ distils over. Sulphuric acid is dibasic,

so the acid consumed is $\frac{0.024}{2}=0.012$ mol. From a 1 M solution this needs $V=\frac{0.012}{1}=0.012$ L, that is 12 mL of 1 M ${\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4$.