Some Basic Concepts of Chemistry Questions — JEE Chemistry

Q1

Thermal decomposition of a Mn compound (X) at 513 K results in compound Y, MnO2 and gaseous product. MnO2 reacts with NaCl and concentrated H2O4 to give a pungent gas Z. X, Y and Z, respectively, are :

A KMnO4, K2MnO4 and Cl2

B K2MnO4, KMnO4 and SO2

C K3MnO4, K2MnO4 and Cl2

D K2MnO4, KMnO4 and Cl2

Correct Answer

Option A

Solution

KMnO4

\overset{{513\,\,K}}\longrightarrow

K2MnO4 + MnO2 + O2 (X) (Y) .tg .tg MnO2 + NaCl + conc H2SO4 $\to$ MnSO4 + NaHSO4 + H2O + Cl2 (Z)

Q2

If a substance '

A

' dissolves in solution of a mixture of '

B

' and '

C

' with their respective number of moles as

\mathrm{n}_{\mathrm{A}}, \mathrm{n}_{\mathrm{B}}

and

\mathrm{n}_{\mathrm{C}_3}

. Mole fraction of

\mathrm{C}

in the solution is

A

\dfrac{n_C}{n_A \times n_B \times n_C}

B

\dfrac{n_B}{n_A+n_B}

C

\dfrac{n_C}{n_A+n_B+n_C}

D

\dfrac{n_C}{n_A-n_B-n_C}

Correct Answer

Option C

Solution

Mole fraction of

\mathrm{C=\frac{n_C}{n_A+n_B+n_C}}

Q3

An organic compound is estimated through Duma's method and was found to evolve 6 moles of CO2. 4 moles of H2O and 1 mole of nitrogen gas. The formula of the compound is :

A C6H8N

B C6H8N2

C C12H8N

D C12H8N2

Correct Answer

Option B

Solution

Molar ratio of C : H : N : : 6 : 8 : 2 i.e., 3 : 4 : 1 Thus, the correct formula is C6H8N2. .

Q4

A solution of sodium sulphate contains 92 g of Na+ ions per kilogram of water. The molality of Na+ ions in that solution in mol kg

-

1 is :

A 12

B 4

C 8

D 16

Correct Answer

Option B

Solution

Molality of Na+ =

{{Moles\,\,of\,\,N{a^ + }} \over {mass\,\,of\,\,{H_2}O\,\,in\,\,kg}}

Moles of Na+ =

{{92} \over {23}}

= 4 Mass of H2O = 1 kg $\therefore$ Molality =

{4 \over 1}

= 4

Q5

The mole fraction of a solvent in aqueous solution of a solute is 0.8. The molality (in mol kg–1 ) of the aqueous solution is :

A 13.88 × 10–1

B 13.88 × 10–3

C 13.88

D 13.88 × 10–2

Correct Answer

Option C

Solution

We know, molality(m) =

{{{x_{solute}}} \over {{x_{solvent}}}} \times {{1000} \over {molar\,weight\,of\,solvent}}

Here solvent is H2O xsolvent = 0.8 $\therefore$ xsolute = 0.2 $\therefore$ Molality =

{{0.2} \over {0.8}} \times {{1000} \over {18}}

= 13.88

Q6

Production of iron in blast furnace follows the following equation Fe3O4(s) + 4CO(g)

\to

3Fe(l) + 4CO2(g) when 4.640 kg of Fe3O4 and 2.520 kg of CO are allowed to react then the amount of iron (in g) produced is : [Given : Molar Atomic mass (g mol

-

1) : Fe = 56, Molar Atomic mass (g mol

-

1) : O = 16, Molar Atomic mass (g mol

-

1) : C = 12]

A 1400

B 2200

C 3360

D 4200

Correct Answer

Option C

Solution

Given, Mass of

F{e_3}{O_4}

= 4.640 kg = 4640 gm Molar mass of

F{e_3}{O_4}

= 56 $\times$ 3 + 16 $\times$ 4 = 232 g $\therefore$ Moles of

F{e_3}{O_4} = {{4640} \over {232}} = 20

Also, given Mass of CO = 2.520 kg = 2520 gm Molar mass of CO = 12 + 16 = 28 gm $\therefore$ Molar of

CO = {{2520} \over {28}} = 90

Here Fe3O4 is limiting reagent as to find limiting reagent, divide the given moles of reactants with their respective stoichiometric coefficient and reactant for which this ratio is minimum will be limiting reagent For

F{e_3}{O_4},{{moles} \over {stoichiometric\,coefficient}} = {{20} \over 1}

For

CO,{{moles} \over {stoichiometric\,coefficient}} = {{90} \over 4} = 22.5

$\therefore$ Fe3O4 is limiting reagent.

Now produced Fe = 20 $\times$ 3 = 60 mol $\therefore$ Weight of Fe = 60 $\times$ 56 = 3360 g

Q7

'25 volume' hydrogen peroxide means

A 1 L marketed solution contains 75 g of H

_2

O

_2

.

B 1 L marketed solution contains 250 g of H

_2

O

_2

.

C 1 L marketed solution contains 25 g of H

_2

O

_2

.

D 100 mL marketed solution contains 25 g of H

_2

O

_2

.

Correct Answer

Option A

Solution

Molarity of $\mathrm{H}_{2} \mathrm{O}_{2}$ soln $=\dfrac{\text{ volume strength }}{11.2}$ $=\dfrac{25}{11.2}=2.23 \mathrm{M}$ $\therefore$ amount of $\mathrm{H}_{2} \mathrm{O}_{2}$ in one litre $=2.23 \times 34=75 \mathrm{gm}$

Q8

6.02

\times

1020 molecules of urea are present in 100 ml of its solution. The concentration of urea solution is (Avogadro constant, NA = 6.02

\times

1023 mol-1)

A 0.02 M

B 0.01 M

C 0.001 M

D 0.1 M

Correct Answer

Option B

Solution

Moles (n) =

{{molecules} \over {{N_A}}}

$\Rightarrow$

{{6.02 \times {{10}^{20}}} \over {6.02 \times {{10}^{23}}}}

$\Rightarrow$ 0.001 Concentration (M) =

{{moles} \over {volume}}

Given volume of urea is 100 ml = 0.1 litre [Note : While calculating Concentration volume should always be in litre.]

$\Rightarrow$

{{0.001} \over {0.1}}

$\Rightarrow$ 0.01 M

Q9

What quantity (in mL) of a 45% acid solution of a mono-protic strong acid must be mixed with a 20% solution of the same acid to produce 800 mL of a 29.875% acid solution ?

A 320

B 325

C 316

D 330

Correct Answer

Option C

Solution

Let the volume of a monoprotic acid solution in mL be V

{{V \times 45} \over {100}} + {{(800 - V)20} \over {100}} = {{800 \times 29.875} \over {100}}

{{9V} \over {20}} + 160 - {V \over 5} = 239

{{5V} \over {20}} = 79 \Rightarrow V = 316

mL

Q10

The molecular formula of a commercial resin used for exchanging ions in water softening is C8H7SO3Na (Mol. Wt. 206). What would be the maximum uptake of Ca2+ ions by the resin when expressed in mole per gram resin?

A 2/309

B 1/412

C 1/103

D 1/206

Correct Answer

Option B

Solution

2

mole of water softner require

1

mole of

C{a^{2 + }}\,\,

ion So,

1

mole of water softner require

{1 \over 2}

mole of

C{a^{2 + }}\,

ion Thus,

{1 \over {2 \times 206}} = {1 \over {412}}mol/g\,\,\,

will be maximum uptake