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)`

Answer

A, C, D

Explanation

i. a. `(Mw of X_(4) O_(6))/("Atomic Weight of" 4X) = ("Weight of" X_(4) O_(6))/("Weight of"X)`
`(4 x + 96)/(4x) = (10)/(5.72)`
`x = 32`
ii.a. Use the formula
`109 = (100 + (18)/(80))%`
`x = 40%` free `SO_(3)`.
Mass of `H_(2) SO_(4) = 100 - 40 = 60 g`
Moles of `SO_(3) = (40)/(80) = 1//2 (q)`
Moles of `H_(2) SO_(4) = (60)/(98) (p)`
`( :. (p - q)/(p + q)) implies ((60)/(98) - (1)/(2))/((60)/(98) + (12)/(2)) ~~ (1)/(9)`
iii.a. (`n` factor for `O_(2) = 4`)
volume strength of `H_(2) O_(2) = (24)/(4) = 6 L`
`6 'VS' of H_(2) O_(2) = 1 N H_(2) O_(2)`
`6 'VS' of H_(2) O_(2) xx vol of H_(2) O_(2) = Vol of O_(2)`
`'VS of H_(2) O_(2) xx 0.1 cm^(3) = 3 dm^(3)`
`'VS' of H_(2) H_(2)O_(2) xx 0.1 dm^(3) = 3 dm^(3)`
`'VS' of H_(2)O_(2) = 30 dm^(3) "or" L`
`6 'VS' of H_(2) O_(2) = 1 N`
`30 'VS' of H_(2) O_(2) = (1 xx 30)/(6) = N`
`M = 5//2 = 2.5`
iv.d. Moles of `S = (4)/(32) = (1)/(8) = 0.125 mol`
`S + O_(2) rarr SO_(2)`
`SO_(2) + Cl_(2) rarr 2Cl^(ɵ) + SO_(4)^(2-)`
`SO_(4)^(2-) + BaSl_(2) rarr BaSO_(4) + 2 Cl^(ɵ)`
Moles of `S =` Moles of `SO_(2) =` Moles of `SO_(4)^(2-) =` Moles of `BaSO_(4) = 0.125 "mol"`.
v.d. `H_(2) + (1)/(2) O_(2) rarr H_(2) O`
Initial mol ratio of `H_(2)//O_(2) = 2.1`
Mol ratio required is the reaction `= 3//1.5`
`= 2 : 1`
Since the mol ratio is name, sok none is the limiting reagent.
vi.a. `{:(Hg^(2+) -=, 4I^() =, HgI_(4)^(2-)),(1 mol -=, 4 mol -=, 1 mol),((1)/(4) mol -=, (4)/(4) mol =, (1)/(4) mol):}`
`= 0.25 mol`