i. `0.21` gm of but-3-yn-2-ol is treated with excess of `C_(2)H_(5)`MgBr at standard condition. The volume of gas evolved is:
(a) `134.4` ml, (b) `146.4` ml
(c) `67.2` ml, (d) `73.2` ml
ii. `0.46` gm of a compound with molecular mass of 92 gm gave 336 ml of a gas at STP when treated with excess of `CH_(3)`MgI. The number of moles in the compound is:
(a) `0.1`, (b) 2
(c) 3, (d) 4
iii. The treatement of `CH_(3)OH` with `CH_(3)Mgl` releases `1.04 ml` of a gas at STP. The mass of `CH_(3)OH` added is:
(a) `1.49` mg,
(b) `2.98` mg
(c) `3.71` mg, (d) `4047` mg
iv. The addition of `4.12` mg of an unknown alcohol, ROH, to `CH_(3)Mgl` releases `1.56` ml of a gas at STP. The molar mass of alcohol is:
(a) 32 gm `mol^(-1)` , (b) 46 gm `mol^(-1)`
(c) `59 gm mol^(-1)` , (d) 74 gm `mol^(-1)`
v. Teh sample of `1.79` mg of a compound of molar mass 90 gm `mol^(-1)` when treated with `CH_(3)Mgl` releases `1.34` mol of a gas at STP. The number of moles of active hydrogen in the molecule is:
(a) 1 , (b) 2
(c) 3 , (d) 4

Explanation

At STP, volume of 1 mol of a gas `=22.4` litres (when P = 1 atm, T = 273 K). If P = 1 bar and T = 273 K, then volume = `22.7` litres. Similary, at standard condition, when P = 1 atm and T = 298 K, then volume = `24.4` litres. When P = 1 bar and T = 298 K, then volume = `24.7` litres.
i.
One mole of compound, i.e., 70 gm gives 2 mol of `C_(2)H_(6)` gas. Seventy grams of compound `implies 2xx24.4` litres of `C_(2)H_(6)` gas at standard condition.
`:. 0.21` gm of the compound `=(2xx2.24xx0.21)/(70)`
`=0.1464` litre `=146.4` ml
ii. `0.46` gm of a compound `implies 336` ml of a gas at STP.
`:. 92` gm of a compound `implies (336xx92)/(0.46) =67200` ml
`=(67200)/(22400) =3` mol
Therefore, number of active H atoms = 3.
iii. a. `CH_(3)OH` contains only one active H atom.
So, 1 mol of `CH_(3)OH` gives 1 mol of `CH_(4)` gas. (molecular weight of MeOH = 32 gm)
22400 ml of `CH_(4)` gas is obtained from 32 gm of MeOH
`:. 1.04` ml of `CH_(4)` gas is obtained `implies(32xx1.04)/(22400)`
`implies 0.00148 g implies 1.49 mg`
iv. c. `1.56` ml of a gas is obtained from `4.12xx10^(-3)` gm of ROH. Therefore, 22400 ml of a gas is obtained from
`(4.12xx10^(-3)xx22400)/(4.56) =59.15 gm mol^(-1)`
c. Let 1 mol of compound gives x mol of a gas.
`:. 90` gm of a compound `implies x xx22400` ml of gas
`1.79xx10^(-3)` gm of a compound gives `implies (x xx22400xx10^(-3))/(90)`
`:. (x xx22400xx1.79xx10^(-3))/(90) =1.34`
`:. x =3`