Some Basic Concepts of Chemistry — Page 6 | NEET Chemistry

Q51

Percentage of Se in peroxidase anhydrous enzyme is 0.5% by weight (at. wt. = 78.4) then minimum molecular weight of peroxide anhydrous enzyme is

A 1.568

\times

10 4

B 1.568

\times

10 3

C 15.68

D 2.136

\times

10 4

Correct Answer

Option A

Solution

{{100} \over {0.5}} \times 78.4 = 1.568 \times {10^4}

Q52

Molarity of liquid HCl, if density of solution is 1.17 g/ce is

A 36.5

B 18.25

C 32.05

D 42.10

Correct Answer

Option C

Solution

Molarity =

{{1.17 \times 1000} \over {36.5 \times 1}} = 32.05

Q53

Oxidation numbers of A, B, C are + 2, +5 and

-

2 respectively. Possible formula of compound is

A A 2 (BC 2 ) 2

B A 3 (BC 4 ) 2

C A 2 (BC 3 ) 2

D A 3 (B 2 C) 2

Correct Answer

Option B

Solution

In A 3 (BC 4 ) 2 , (+2) $\times$ 3 + 2[+5+4(-2)] $\Rightarrow$ +6 + 10 -16 = 0 $\therefore$ Oxidation numbers of A, B, C are + 2, +5 and $-$ 2 respectively.

Q54

Volume of CO 2 obtained by the complete decomposition of 9.85 g of BaCO 3 is

A 2.24 L

B 1.118 L

C 0.84 L

D 0.56 L

Correct Answer

Option B

Solution

BaCO 3 $\to$ BaO + CO 2 197.34 g $\to$ 22.4 L at N.T.P 9.85 g $\to$

{{22.4} \over {197.34}} \times 9.85 = 1.118\,L

Q55

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}}

Q56

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

Q57

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

Q58

'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 $=2.23 \times 34=75 \mathrm{gm}$

Q59

The amount of sugar (C12H22O11) required to prepare 2L of its 0.1 M aqueous solution is :

A 17.1 g

B 34.2 g

C 68.4 g

D 136.8 g

Correct Answer

Option C

Solution

Molarity =

{{{{(n)}_{solute}}} \over {{V_{solution}}(in\,\,lit)}}

0.1 =

{{wt./342} \over 2}

wt (C12H22O11) = 68.4 gram

Q60

The percentage composition of carbon by mole in methane is :

A 80%

B 20%

C 75%

D 25%

Correct Answer

Option B

Solution

In CH4, there are 5 atoms (1C + 4H). $\therefore$ % composition of C by mole in CH4 =

{1 \over 5}

$\times$ 100 = 20 %