2. 0.22 g of a colourless oxide of nitrogen occupies 112 ml at STP. The compound is:
a. NO
b. N2O
c. NO2
d. N2O2
Quick Solution
0.22 g of a colourless oxide of nitrogen occupies 112 ml at STP, corresponding to a molar mass of 44 g/mol, which matches N₂O among the options. This identification relies on calculating moles from volume using the molar volume of 22.4 L (22400 ml) at STP, then deriving molar mass as mass divided by moles.[web:9][web:10] The compound is N₂O (option b).[web:6]
Solution Steps
At STP, 1 mole of ideal gas occupies 22400 ml.[web:9] For 112 ml, moles \( n = \frac{112}{22400} = 0.005 \) mol. Molar mass \( M = \frac{0.22}{0.005} = 44 \) g/mol.[web:1]
Option Analysis
| Option | Formula | Molar Mass (g/mol) | Matches 44 g/mol? |
|---|---|---|---|
| a | NO | 30 | No[web:6] |
| b | N₂O | 44 | Yes[web:6] |
| c | NO₂ | 46 | No[web:6] |
| d | N₂O₂ | 60 | No[web:6] |
CSIR NET Exam Context
The question “0.22 g oxide of nitrogen 112 ml STP” tests molar mass determination for 0.22 g colourless oxide of nitrogen occupies 112 ml at STP among NO, N₂O, NO₂, N₂O₂. This CSIR NET-style problem uses STP gas laws where 22400 ml equals 1 mole.[web:10]
Molar Mass Calculation
Moles = volume / 22400 ml = 112 / 22400 = 0.005 mol. Thus, molar mass = 0.22 g / 0.005 mol = 44 g/mol, matching N₂O.[web:6]
Why Other Options Fail
- NO (30 g/mol): Too light for given mass-volume.
- NO₂ (46 g/mol): Slightly heavier.
- N₂O₂ (60 g/mol): Much heavier.[web:6]
N₂O, a colourless gas, fits perfectly for exam prep on nitrogen oxides.[web:3]
