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Question: In Duma's method of estimation of nitrogen, 0.1840 g of an organic compound gave 30 mL of nitrogen c...

In Duma's method of estimation of nitrogen, 0.1840 g of an organic compound gave 30 mL of nitrogen collected at 287 K and 758 mm of Hg pressure. The percentage composition of nitrogen in the compound is ____. (Round off to the Nearest Integer).

[Given: Aqueous tension at 287 K = 14 mm of Hg]

Answer

19

Explanation

Solution

Step 1: Calculate the pressure of dry nitrogen gas.

The total pressure is the sum of the partial pressure of dry nitrogen and the aqueous tension.

Pressure of dry N₂ (P1P_1) = Total pressure - Aqueous tension

P1P_1 = 758 mm Hg - 14 mm Hg = 744 mm Hg

Step 2: Convert the volume of nitrogen gas from the given conditions (P1P_1, V1V_1, T1T_1) to Standard Temperature and Pressure (STP) conditions (P0P_0, V0V_0, T0T_0).

STP conditions are typically defined as T0T_0 = 273 K and P0P_0 = 760 mm Hg.

Using the combined gas law: P1V1T1=P0V0T0\frac{P_1 V_1}{T_1} = \frac{P_0 V_0}{T_0}

V0=P1V1T0P0T1V_0 = \frac{P_1 V_1 T_0}{P_0 T_1}

V0=744 mm Hg×30 mL×273 K760 mm Hg×287 KV_0 = \frac{744 \text{ mm Hg} \times 30 \text{ mL} \times 273 \text{ K}}{760 \text{ mm Hg} \times 287 \text{ K}}

V0=6094800218120 mLV_0 = \frac{6094800}{218120} \text{ mL}

V0≈27.94237 mLV_0 \approx 27.94237 \text{ mL}

Step 3: Calculate the mass of nitrogen corresponding to the volume at STP.

At STP, the molar volume of any ideal gas is 22.4 L or 22400 mL. The molar mass of nitrogen gas (N₂) is 28 g/mol.

Mass of N₂ = (Volume of N₂ at STP / Molar volume at STP) ×\times Molar mass of N₂

Mass of N₂ = V022400 mL/mol×28 g/mol\frac{V_0}{22400 \text{ mL/mol}} \times 28 \text{ g/mol}

Mass of N₂ = 27.94237 mL22400 mL×28 g\frac{27.94237 \text{ mL}}{22400 \text{ mL}} \times 28 \text{ g}

Mass of N₂ ≈0.001247427×28 g\approx 0.001247427 \times 28 \text{ g}

Mass of N₂ ≈0.034928 g\approx 0.034928 \text{ g}

Step 4: Calculate the percentage composition of nitrogen in the organic compound.

Percentage of N = Mass of N2Mass of organic compound×100\frac{\text{Mass of N}_2}{\text{Mass of organic compound}} \times 100

Percentage of N = 0.034928 g0.1840 g×100\frac{0.034928 \text{ g}}{0.1840 \text{ g}} \times 100

Percentage of N ≈0.189826×100\approx 0.189826 \times 100

Percentage of N ≈18.9826%\approx 18.9826 \%

Step 5: Round off the percentage to the Nearest Integer.

18.9826 % rounded to the nearest integer is 19 %.