Some Basic Concepts of Chemistry — Page 2 | NEET Chemistry

Q11

The density of 1 M solution of a compound 'X' is 1.25 g mL

^{-1}

. The correct option for the molality of solution is (Molar mass of compound X = 85 g):

A 0.705 m

B 1.208 m

C 1.165 m

D 0.858 m

Correct Answer

Option D

Solution

\begin{aligned} & \mathrm{m}=\frac{1000 \times \mathrm{M}}{1000 \times \mathrm{d}-\mathrm{MM}_{\mathrm{w}}} \\ & \mathrm{m}=\frac{1000 \times 1}{1000 \times 1.25-1 \times 85} \\ & \mathrm{~m}=\frac{1000}{1165}=0.858\end{aligned}

Q12

The right option for the mass of

\mathrm{CO}_{2}

produced by heating

20 \mathrm{~g}

of

20 \%

pure limestone is (Atomic mass of

\mathrm{Ca}=40

)

\left[\mathrm{CaCO}_{3} \stackrel{1200 \mathrm{~K}}{\longrightarrow} \mathrm{CaO}+\mathrm{CO}_{2}\right]

A 1.76 g

B 2.64 g

C 1.32 g

D 1.12 g

Correct Answer

Option A

Solution

Weight of impure limestone

=20 \mathrm{~g}

Weight of pure limestone

\left(\mathrm{CaCO}_{3}\right)=20 \%

of

20 \mathrm{~g}

=\frac{20}{100} \times 20

=4 \mathrm{~g}

\mathrm{n}_{\mathrm{CaCO}_{3}}=\frac{4}{100}=0.04

Image

\mathrm{n}_{\mathrm{CO}_{2}}=0.04

\mathrm{W}_{\mathrm{CO}_{2}}=0.04 \times 44

=1.76 \mathrm{~g}

Q13

The density of the solution is 2.15 g mL

-

1 , then mass of 2.5 mL solution in correct significant figures is :

A 53.75 g

B 5375

\times

10

-

3 g

C 5.4 g

D 5.38 g

Correct Answer

Option C

Solution

Mass = Volume $\times$ Density = 2.5 $\times$ 2.15 = 5.375 g Since 2.5 has two significant figures, so the mass of solution in correct significant figures will be 5.4 g.

Q14

What fraction of Fe exists as Fe(III) in Fe 0.96 O ? (Consider Fe 0.96 O to be made up of Fe(II) and Fe(III) only)

A

{1 \over {20}}

B

{1 \over {12}}

C 0.08

D

{1 \over {16}}

Correct Answer

Option B

Solution

Fe 0.96 O Let Fe(II) present in Fe 0.96 O = x Fe(III) present = (0.96 $-$ x) Total charge on Fe = 2x + (0.96 $-$ x)3 Total charge on O = $-$ 2 2x + (0.96 $-$ x)3 = 2 2x + 2.88 $-$ 3x = 2 $-$ x = $-$ 0.88 x = 0.88 Fe 2+ = 0.88, Fe 3+ = 0.08 Fraction of Fe 3+ =

{{0.08} \over {0.96}} = {1 \over {12}}

Q15

In one molal solution that contains 0.5 mole of a solute, there is

A 500 mL of solvent

B 500 g of solvent

C 100 mL of solvent

D 1000 g of solvent

Correct Answer

Option B

Solution

Molality is the moles of solute dissolved per kg of solvent therefore 500 g, 1 molal solution contains 0.5 of solute, as

m = {{Moles\,of\,solute} \over {Mass\,of\,solvent\,(in\,kg)}}

1 = {{0.5} \over {Mass\,of\,solvent\,(in\,kg)}}

$\therefore$ Mass of solvent (in kg) = 0.5 = 500 g

Q16

What mass of 95% pure CaCO 3 will be required to neutralise 50 mL of 0.5 M HCl solution according to the following reaction? CaCO 3(s) + 2HCl (aq)

\to

CaCl 2(aq) + Co 2(g) + 2H 2 O (l) [Calculate upto second place of decimal point]

A 1.25 g

B 1.32 g

C 3.65 g

D 9.50 g

Correct Answer

Option B

Solution

Let m gram mass of CaCO 3 is required Pure CaCO 3 in m gram

= {{95} \over {100}} \times m

Moles of CaCO 3

= {{95} \over {100}} \times {m \over {100}}

Moles of HCl required = 2 $\times$ moles of CaCO 3

= 2 \times {{95} \over {100}} \times {m \over {100}}

2 \times {{95} \over {100}} \times {m \over {100}} = {{50} \over {1000}} \times 0.5

m = 1.315\,g \approx 1.32\,g

Q17

An organic compound contains 78% (by wt.) carbon and remaining percentage of hydrogen. The right option for the empirical formula of this compound is : [Atomic wt. of C is 12, H is 1]

A CH 4

B CH

C CH 2

D CH 3

Correct Answer

Option D

Solution

Based on above calculation, possible empirical formula is CH 3 .

Q18

Which one of the following has maximum number of atoms?

A 1 g of Mg(s) [Atomic mass of Mg = 24]

B 1 g of O 2 (g) [Atomic mass of O = 24]

C 1 g of Li(s) [Atomic mass of Li = 7]

D 1 g of Ag(s) [Atomic mass of Ag = 108]

Correct Answer

Option C

Solution

Number of atoms =

{w \over {molarmass}} \times {N_A} \times Atomicity

Number of Mg atoms =

{1 \over {24}} \times {N_A}

Number of O atoms =

{1 \over {32}} \times 2 \times {N_A}

Number of Li atoms =

{1 \over 7} \times {N_A}

Number of Ag atoms =

{1 \over {108}} \times {N_A}

Q19

The number of protons, neutrons and electrons in 175 71 Lu, respectively, are :

A 104, 71 and 71

B 71, 71 and 104

C 175, 104 and 71

D 71, 104 and 71

Correct Answer

Option D

Solution

No. of Protons = 71 = No. of Electrons No. of Neutrons = Mass No. $-$ No. of Protons = 175 $-$ 71 = 104

Q20

The number of moles of hydrogen molecules required to produce 20 moles of ammonia through Haber’s process is :

A 10

B 20

C 30

D 40

Correct Answer

Option C

Solution

According to Haber's process, N 2 (g) + 3H 2 (g) ⇌ 2NH 3 (g) 2 moles of NH 3 uses 3 moles of H 2 So, 20 moles of NH 3 will use

{3 \over 2}

$\times$ 20 moles of H 2 = 30 moles of H 2