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∴
4 g of NH 4 Cl will be decomposed by
74
107 × 4 g = 2.766 g of Ca(OH) 2
Thus, the mass of slaked lime required = 2.766 g
31. (c) :
NaBH 4
CH3CH2C
CCH2CHO
Hex-3-ynal
PBr 3
CH3CH2C
CCH2CHOH
2
CH3CH2C
CCH2CHBr
2
( P)
+ (ii) H 3 O
Mg
ether
CH3CH2C
CCH2CHMgBr
2
SOCl
CH3CH2C
2
CCH2CHCOOH
2
( Q)
quinoline
CH3CH2C
CCH2CHCOCl
2
H 2/Pb-BaSO4
HC
CH
CH 3 CH 2 CH2CH2CHO
( S)
32. (b): The action of heat on boric acid is shown as :
375 K
435 K
4HBO
3 3 →
4HBO
4 2 →HBO
− HO 2
−HO
2 4 7
Orthoboric acid
Metabo ric 2 Tetraboric
or
(Boric acid)
acid
acid
redheat
→2 BO
−HO
2 3
2 Boron
trioxide
33. (d): MnO – 4 is the strongest oxidising agent because
it has the highest reduction potential value.
Heat
34. (b): Fe2( SO4) 3 →Fe2O3+
3SO3
35. (d): (+)–Lactose is a reducing sugar and shows
mutarotation.
36. (c) : HA + B BH + + A – ; K = 100
K f = 10 5 K 5
f 10
, K b = ?, Kb
= = =
K 100 10 3
37. (b): For a gaseous mixture of C 2 H 6 and C 2 H 4 ,
PV = nRT
∴ 1 × 40 = n × 0.082 × 400 ⇒ n = 1.2195
∴ Total moles of C 2 H 6 and C 2 H 4 = 1.2195
Let number of moles of C 2 H 6 and C 2 H 4 be a and b
respectively.
a + b = 1.2195
...(i)
C 2 H 6 + 7/2O 2 2CO 2 + 3H 2 O
C 2 H 4 + 3O 2 2CO 2 + 2H 2 O
∴ Number of moles of O 2 needed for complete
reaction of the mixture
7a
= + 3b = 130
2 32
...(ii)
Solving eqs. (i) and (ii), we get, a = 0.808; b = 0.4115
∴ Mole fraction of C 2 H 6 = 0.808/1.2195 = 0.66
and mole fraction of C 2 H 4 = 0 . 4115 = 0.34
1.
2195
38. (d): n X = n Y = 1 or n X = 1
nY
1 1 1 1
xX
= = , xY
= =
1 + 1 2 1 + 1 2
P = p° X × x X + p° Y × x Y = 400 mm
or
1 1
p°+ X p°= Y 400 mm
2 2
...(i)
n′
When X 1
= at the same temperature,
nY′
2
x′ X
= 1 3 and x′ Y = 2 3
∴ P′ = p° X
× x′ X
+ p° Y
× x′ Y
= 350 mm
or
1 2
p°+ p° = 350 mm
3 X 3 Y
...(ii)
Solving equations (i) and (ii), we get, p° X = 550 mm,
p° Y = 250 mm
39. (a) : Li 2 O reacts with CO 2 as :
Li 2 O + CO 2 Li 2 CO 3
i.e., 1 mole of Li 2 O (= 30 g Li 2 O) reacts with 22.4 L
of CO 2 at STP
22. 4×
1000
or 1000 g Li 2 O absorbs =
30
= 746.66 L of CO 2
∴ Absorption efficiency is 746.66 L/kg
40. (a) : Due to inert pair effect, heavier p-block
elements show low (two units less) oxidation state
as the most stable one.
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CHEMISTRY TODAY | APRIL ‘17 17