1
MCQ (Single Correct Answer)

JEE Main 2016 (Online) 9th April Morning Slot

The amount of arsenic pentasulphide that can be obtained when 35.5 g arsenic acid istreated with excess H2S in the presence of conc. HCl ( assuming 100% conversion) is :
A
0.50 mol
B
0.25 mol
C
0.125 mol
D
0.333 mol

Explanation

2H3 As O4 + 5H2S  $$\mathrel{\mathop{\kern0pt\longrightarrow} \limits_{HCl}^{conc}} $$   As2S5 + 8H2O

35.5 g of H3AsO4 = $${{35.5} \over {142}}$$ = 0.25 moles

Let, As2S5 produced = n moles.

$$\therefore\,\,\,$$ $${{0.25} \over 2} = {n \over 1}$$

$$ \Rightarrow $$$$\,\,\,$$ n = 0.125 mol.
2
MCQ (Single Correct Answer)

JEE Main 2016 (Online) 9th April Morning Slot

An organic compound contains C, H and S. The minimum molecular weight of the compound containing 8% sulphur is :

(atomic weight of S = 32 amu)
A
200 g mol$$-$$1
B
400 g mol$$-$$1
C
600 g mol−1
D
300 g mol−1
3
MCQ (Single Correct Answer)

JEE Main 2016 (Online) 9th April Morning Slot

5 L of an alkane requires 25 L of oxygen for its complete combustion. If all volumes are measured at constant temperature and pressure, the alkane is :
A
Ethane
B
Propane
C
Butane
D
Isobutane

Explanation

We know, conbustion of Hydrocarbon

Cx Hy + (x + $${y \over 4}$$)O2  $$ \to $$  x CO2 + $${y \over 2}$$ H2O

So, to consume 1 mole of Cx Hy we need

(x + $${y \over 4}$$) mole of O2 gas.

As both Cx Hy and O2 are gas,

So avogadro's law is applicable on them.

At constant temperature and pressure according to avogadro's law, volume $$ \propto $$ mole

So, for 5L of Cx Hy the

volume of O2 = 5(x + $${y \over 4}$$)

According to the question,

5(x + $${y \over 4}$$) = 25

$$\therefore\,\,\,$$ (x + $${y \over 4}$$) = 5 . . . . . (1)

For Ethane (C2 H6), x = 2 and y = 6

$$\therefore\,\,\,$$ x + $${y \over 4}$$ = 2 + $${6 \over 4}$$ $$ \ne $$ 5

$$\therefore\,\,\,$$ C2 H6 can't be the required alkane.

For Propane (C3 H8), x = 3. and y = 8

$$\therefore\,\,\,$$ x + $${y \over 4}$$ = 3 + $${8 \over 4}$$ = 5

$$\therefore\,\,\,$$ Propane (C3 H8) is the right alkane.

Similarly for Butane (C4 H10) and Isobutane (C4 H10) you can check
x + $${y \over 4}$$ $$ \ne $$ 5.
4
MCQ (Single Correct Answer)

JEE Main 2016 (Online) 10th April Morning Slot

The volume of 0.1N dibasic acid sufficient to neutralize 1 g of a base that furnishes 0.04 mole of OH in aqueous solution is :
A
200 mL
B
400 mL
C
600 mL
D
800 mL

Explanation

According to law of equivalence,

Equivalence of acid = Equivalence of base.

Equivalence of acid = Normality x volume = 0.1 $$ \times $$ v

As we know base produce OH$$-$$ ion, so moles of base is same as moles of OH$$-$$ ion = 0.04

Another formula of equivalence = n factor $$ \times $$ number of moles

$$\therefore\,\,\,$$ Equivalance of base = n factor of OH$$-$$ $$ \times $$ moles of OH$$-$$ = 1 $$ \times $$ 0.04

As for any ion, the charge of that ion is the n factor of that ion. Here OH$$-$$ has 1 negative charge so it's n factor = 1

$$\therefore\,\,\,$$ 0.1 $$ \times $$ v = 1 $$ \times $$ 0.04

$$ \Rightarrow $$$$\,\,\,$$ v = 0.4 L

=   0.4 $$ \times $$ 1000

=    400 ml.

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