Question

Given below are some nitrogen containing compounds:
Each of them is treated with HCl separately. 1.0 g of the most basic compound will consume ...... mg of HCl.
(Given Molar mass in g mol\(^{-1}\): C = 12, H = 1, O = 16, Cl = 35.5.)

Answer 1

Correct Answer: 341

To address this problem, we must first ascertain the most fundamental compound from the provided choices. Subsequently, we will compute the quantity of HCl that 1.0 g of this identified compound would consume.

1. Identification of the Most Basic Compound:
A comparative analysis of the basicity of 4-nitroaniline, benzylamine, N-phenylacetamide, and aniline is required. Basicity is contingent upon the accessibility of the nitrogen lone pair for protonation.

2. Basicity Assessment:
4-Nitroaniline: The nitro group, being electron-withdrawing, diminishes basicity.
Benzylamine: The nitrogen atom is bonded to a benzyl group, rendering it more basic than aniline.
N-Phenylacetamide: This is an amide, exhibiting very weak basicity due to resonance with the carbonyl group.
Aniline: The lone pair is delocalized within the benzene ring, reducing basicity.
Consequently, benzylamine is identified as the most basic compound.

3. Calculation of Benzylamine Molar Mass ($C_7H_9N$):
Molar mass = (7 × 12) + (9 × 1) + (1 × 14) = 84 + 9 + 14 = 107 g/mol

4. Calculation of Benzylamine Moles:
Given 1.0 g of benzylamine, the number of moles is calculated as: moles = mass / molar mass = 1.0 g / 107 g/mol = 1/107 mol

5. Reaction with HCl:
Benzylamine reacts with HCl in a stoichiometric ratio of 1:1: $C_7H_9N + HCl \rightarrow C_7H_9NH^+Cl^-$.
Therefore, 1/107 moles of benzylamine react with 1/107 moles of HCl.

6. Calculation of HCl Molar Mass:
Molar mass = 1 (for H) + 35.5 (for Cl) = 36.5 g/mol

7. Calculation of HCl Mass Consumed:
Mass of HCl = moles × molar mass = (1/107 mol) × (36.5 g/mol) = 36.5 / 107 g

8. Conversion to Milligrams:
Mass of HCl (in mg) = (36.5 / 107) × 1000 mg ≈ 0.3411 × 1000 mg ≈ 341.1 mg

Conclusion:
1.0 g of benzylamine will consume approximately 341.1 mg of HCl.