During estimation of Nitrogen by Dumas' method of compound X (0.42 g) :
mL of $ N_2 $ gas will be liberated at STP. (nearest integer) $\text{(Given molar mass in g mol}^{-1}\text{ : C : 12, H : 1, N : 14})$
During estimation of Nitrogen by Dumas' method of compound X (0.42 g) :
mL of $ N_2 $ gas will be liberated at STP. (nearest integer) $\text{(Given molar mass in g mol}^{-1}\text{ : C : 12, H : 1, N : 14})$
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Correct Answer: 109
Approach Solution - 1
To estimate the volume of $N_2$ gas liberated during the Dumas' method at STP, we start by identifying the compound. The structure given corresponds to piperazine, which has the molecular formula $C_4H_{10}N_2$.
Molecular Weight Calculation:
Molecular weight of piperazine $=4 \times 12\ (\text{C}) + 10 \times 1\ (\text{H}) + 2 \times 14\ (\text{N}) = 48 + 10 + 28 = 86\ \text{g/mol}$
Calculating Moles of Compound:
Given mass of compound $X = 0.42\ \text{g}$
Moles of compound $=\frac{0.42}{86}\ \text{mol} \approx 0.0048845\ \text{mol}$
Moles and Volume of $N_2$ Gas:
Each molecule of piperazine releases 1 molecule of $N_2$.
Moles of $N_2 = 0.0048845\ \text{mol}$
Volume of $N_2$ at STP $= 0.0048845\ \text{mol} \times 22.4\ \text{L/mol} = 0.109\ \text{L} = 109\ \text{mL}$
Conclusion:
The volume of $N_2$ gas liberated is 109 mL, which is within the given range of 109,109.
Approach Solution -2
To find the volume of $N_2$ gas liberated when 0.42 g of compound X is used in Dumas' method, we first identify the compound's formula. From the structure, compound X is piperazine, C4H10N2, with two nitrogen atoms per molecule. The molar mass of C4H10N2 is calculated as follows:
- Carbon (C): 4 × 12 = 48 g/mol
- Hydrogen (H): 10 × 1 = 10 g/mol
- Nitrogen (N): 2 × 14 = 28 g/mol
Total molar mass = 48 + 10 + 28 = 86 g/mol
Given 0.42 g of compound X, the moles of X are:
0.42 g / 86 g/mol = 0.00488 mol
Since each molecule of C4H10N2 contains 2 nitrogen atoms, the moles of $N_2$ gas produced are:
0.00488 mol of X × (1 mol $N_2$ / 1 mol X) = 0.00488 mol $N_2$
At STP (273 K, 1 atm), 1 mole of any gas occupies 22.4 L. Thus, the volume of $N_2$ liberated is:
0.00488 mol × 22.4 L/mol = 0.109312 L
Converting to mL, we have 0.109312 × 1000 = 109.312 mL.
Conclusion: The volume of $N_2$ gas liberated at STP is 109 mL, which matches the expected range (109,109).
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