0.53 g of an organic compound (x) when heated with excess of nitric acid (concentrated) and then with silver nitrate gave 0.75 g of silver bromide precipitate. 1.0 g of (x) gave 1.32 g of CO₂ gas on combustion. The percentage of hydrogen in the compound (x) is ____ %. (Nearest Integer) [Given: Molar mass in gmol⁻¹ H : 1, C : 12, Br : 80, Ag : 108, O ; Compound (x) : CₓHᵧBr

0.53 g of an organic compound when treated with nitric acid and silver nitrate gives AgBr and CO2 on combustion. Find percentage of hydrogen.

Q. 0.53 g of an organic compound (x) when heated with excess of nitric acid (concentrated) and then with silver nitrate gave 0.75 g of silver bromide precipitate. 1.0 g of (x) gave 1.32 g of CO₂ gas on combustion. The percentage of hydrogen in the compound (x) is ____ %. (Nearest Integer)

[Given: Molar mass in gmol⁻¹ H : 1, C : 12, Br : 80, Ag : 108, O ; Compound (x) : C_xH_yBr_z]

Correct Answer: 4

Explanation

Step 1: Calculate amount of bromine present

Molar mass of AgBr = 108 + 80 = 188 g mol⁻¹

Moles of AgBr = 0.75 / 188 = 0.00399 mol

Each mole of AgBr contains 1 mole of Br.

Mass of Br = 0.00399 × 80 = 0.319 g

Step 2: Calculate percentage of bromine

% Br = (0.319 / 0.53) × 100 ≈ 60.2%

Step 3: Calculate amount of carbon from CO₂

Moles of CO₂ = 1.32 / 44 = 0.03 mol

Moles of C = 0.03 mol

Mass of C = 0.03 × 12 = 0.36 g

Step 4: Calculate percentage of carbon

% C = (0.36 / 1.0) × 100 = 36%

Step 5: Calculate percentage of hydrogen

% H = 100 − (36 + 60.2) ≈ 3.8%

Nearest integer value = 4%

Explanation

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