X g of nitrobenzene on nitration gave 4.2 g of m-dinitrobenzene. X = g. (nearest integer) [Given : molar mass (in g mol-1) C : 12, H : 1, O : 16, N : 14]
Answer
3
Solution
C6H5NO2 conc. HNO3/conc. H2SO4 m-C6H4(NO2)2
1 mol of nitrobenzene gives 1 mol of m-dinitrobenzene.
Molar mass of C6H5NO2 = 72 + 5 + 14 + 32 = 123 g mol-1
Molar mass of C6H4(NO2)2 = 72 + 4 + 28 + 64 = 168 g mol-1
moles of m-dinitrobenzene = 4.2168 = 0.025 mol
moles of nitrobenzene needed = 0.025 mol
X = 0.025 123 = 3.075 g 3 g
Answer: 3