Objective question .
i. A certains compound has the molecular formula `X_(4) O_(6)`. If `10 g of X_(4) O_(6)` has `5.72 g X`, then atomic mass of `X` is:
a. 32 amu b. 42 amu c. 98 amu d. 37 amu
ii. For 109% labelled oleum, if the number of moles of `H_(2)SO_(4)` and free `SO_(3)` be `p` and `q`, respectively, then what will be the value of `(p - q)/(p + q)`
a. `1//9` b. 9 c. 18 d. `1//3`
iii. Hydrogen peroxide in aqueous solution decomposes on warming to give oxygen according to the equation,
`2H_(2) O_(2) (aq) rarr 2 H_(2) O (l) + O_(2) (g)`
Under conditions where 1 mol gas occupies `24 dm^(3), 100 cm^(3)` of `X M` solution of `H_(2) O_(2)` produces `3 dm^(3)` of `O_(2)`. Thus, `X` is
a. 2.5 b. 0.5 c. 0.25 d. 1
iv. `4 g` of sulphur is burnt to form `SO_(2)` which is oxidised by `Cl_(2)` water. The solution is then treated with `BaCl_(2)` solution. The amount of `BaSO_(4)` precipitated is:
a. 0.24 mol b. 0.5 mol
c. 1 mol d. 0.125 mol
v. A reaction occurs between 3 moles of `H_(2)` and 1.5 moles of `O_(2)` to give some amount of `H_(2) O`. The limiting reagent in this reaction is
a. `H_(2)` and `O_(2)` both b. `O_(2)`
c. `H_(2)` d. Neither of them
vi. `4 I^(ɵ) + Hg^(2+) rarr HgO_(4)^(-)`, 1 mole each of `Hg^(2+)` and `I^(ɵ)` will form:
a. 1 mol of `HgI_(4)^(2-)`
b. 0.5 mol of `HgI_(4)^(-2)`
0.25 mol of `HgI_(4)^(2-)`
2 mol of `HgI_(4)^(-2)`

Explanation

Let's solve the questions step by step:

### Question i: Atomic Mass of X in X₄O₆

1. **Given Data**:
- Molecular formula: X₄O₆
- Mass of X₄O₆ = 10 g
- Mass of X = 5.72 g

2. **Calculate Mass of Oxygen**:
- Mass of O = Mass of X₄O₆ - Mass of X
- Mass of O = 10 g - 5.72 g = 4.28 g

3. **Calculate Moles of X and O**:
- Moles of X = mass of X / atomic mass of X = 5.72 / Mₓ
- Moles of O = mass of O / atomic mass of O = 4.28 / 16 (since atomic mass of O = 16)

4. **Using the Molecular Formula**:
- From the formula X₄O₆, the ratio of moles of X to moles of O is 4:6 or 2:3.
- Therefore, (5.72 / Mₓ) / (4.28 / 16) = 2/3

5. **Cross-multiplying**:
- 5.72 * 3 * 16 = 4.28 * 2 * Mₓ
- 274.56 = 8.56 * Mₓ

6. **Solving for Mₓ**:
- Mₓ = 274.56 / 8.56 = 32 amu

**Final Answer**: The atomic mass of X is **32 amu** (Option a).