Some Basic Concepts of Chemistry — Page 3 | JEE Chemistry

Q21

The ratio of mass percent of C and H of an organic compound (CXHYOZ) is 6 : 1. If one molecule of the above compound (CXHYOZ) contains half as much oxygen as required to burn one molecule of compound CXHY completely to CO2 and H2O. The empirical formula of compound CXHYOZ is

A C2H4O3

B C3H6O3

C C2H4O

D C3H4O2

Correct Answer

Option A

Solution

\therefore\,\,\,

The ratio of no of atoms of C and H in one molecule of Cx Hy Oz = 1 : 2

\therefore\,\,\,

y = 2x In one molecule of Cx Hy Oz compound contain z atoms of oxygen.

According to the question, no of atoms of oxygen required to burn CxHy completely should be twice of z atoms of oxygen.

CxHy + (x +

{y \over 4}

) O2 $\to$ HCO2 +

{y \over 2}

H2O No. of O2 molecules required = (x +

{y \over 4}

)

\therefore\,\,\,

No. of O atoms required = 2 (x +

{y \over 4}

) According to question,

2\left( {x + {y \over 4}} \right) = 2z

\Rightarrow \,\,\,2\left( {x + {{2x} \over 4}} \right) = 2z\,\,\,

[ Putting y = 2x ]

\Rightarrow \,\,\,\,{{3x} \over 2} = z

\therefore\,\,\,

x : y : z = x : 2x :

{{3x} \over 2}

= 2 : 4 : 3

\therefore\,\,\,

Empirical formula = C2H4O3

Q22

The ratio of masses of oxygen and nitrogen in a particular gaseous mixture is 1 : 4. The ratio of number of their molecule is:

A 1: 8

B 3 : 16

C 1 : 4

D 7 : 32

Correct Answer

Option D

Solution

Given ratio of mass of O2 and N2 = 1 : 4 Let mass of O2 = w and mass of N2 = 4w $\therefore$ Number of moles of O2 =

{w \over {32}}

Number of molecules of O2 =

{w \over {32}}

$\times$ NA Number of moles of N2 =

{4w \over {28}}

Number of molecules of N2 =

{4w \over {28}}

$\times$ NA $\therefore$ Ratio of molecules of O2 and N2 =

{w \over {32}}

$\times$ NA :

{4w \over {28}}

$\times$ NA = 7 : 32

Q23

8 g of NaOH is dissolved in 18g of H2O. Mole fraction of NaOH in solution and molality (in mol kg–1) of the solution respectively are -

A 0.2, 11.11

B 0.167, 22.20

C 0.167, 11.11

D 0.2, 22.20

Correct Answer

Option C

Solution

8 gm of NaOH =

{8 \over {40}}

= 0.2 mol of NaOH 18 gm of H2O =

{18 \over {18}}

= 1 mol of H2O $\therefore$ Total mole = 1 + 0.2 = 1.2 mol $\therefore$ Mole fraction of NaOH =

{{0.2} \over {1.2}}

= 0.167 We know, Molality =

{{Moles\,\,of\,\,solute} \over {Weight\,of\,solvent}} \times

1000 =

{{0.2} \over {18}} \times

1000 = 11.11

Q24

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

Q25

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 %

Q26

A solution of two components containing n1 moles of the 1st component and n2 moles of the 2nd component is prepared. M1 and M2 are the molecular weights of component 1 and 2 respectively. If d is the density of the solution in g mL–1, C2 is the molarity and x2 is the mole fraction of the 2nd component, then C2 can be expressed as :

A

{C_2} = {{1000{x_2}} \over {{M_1} + {x_2}\left( {{M_2} - {M_1}} \right)}}

B

{C_2} = {{1000d{x_2}} \over {{M_1} + {x_2}\left( {{M_2} - {M_1}} \right)}}

C

{C_2} = {{d{x_2}} \over {{M_1} + {x_2}\left( {{M_2} - {M_1}} \right)}}

D

{C_2} = {{d{x_1}} \over {{M_1} + {x_2}\left( {{M_2} - {M_1}} \right)}}

Correct Answer

Option B

Solution

1st component 2nd component Mole n1 n2 Molecular Weight M1 M2 Mass n1M1 n2M2 Mass of solution = n1M1 + n2M2 density of the solution = d C2 = Molarity of 2nd Component x2 = Mole-fraction of 2nd Component x2 =

{{{n_2}} \over {{n_1} + {n_2}}}

$\Rightarrow$ 1 - x2 =

{{{n_1}} \over {{n_1} + {n_2}}}

Molarity of 2nd Component, C2 = n2 Volume of solution(ml) $\times$ 1000 =

{{{n_2}} \over {{{Mass\,of\,Solution(ml)} \over {Density\,of\,Solution(g/ml)}}}} \times 1000

=

{{d \times {n_2} \times 1000} \over {{n_1}{M_1} + {n_2}{M_2}}}

=

{{d \times {{{n_2}} \over {{n_1} + {n_2}}} \times 1000} \over {{{{n_1}} \over {{n_1} + {n_2}}}{M_1} + {{{n_2}} \over {{n_1} + {n_2}}}{M_2}}}

(Dividing componendo and dividendo by n1 + n2) =

{{d \times {x_2} \times 1000} \over {\left[ {\left( {1 - {x_2}} \right){M_1} + {x_2}{M_2}} \right]}}

=

{{1000d{x_2}} \over {{M_1} + {x_2}\left( {{M_2} - {M_1}} \right)}}

Q27

Complete combustion of 1.80 g of an oxygen containing compound (CxHyOz) gave 2.64 g of CO2 and 1.08 g of H2O. The percentage of oxygen in the organic compound is :

A 51.63

B 50.33

C 63.53

D 53.33

Correct Answer

Option D

Solution

CxHyOz + O2 $\to$ xCO2 +

{y \over 2}{H_2}O

2.64 g of CO2 contains 0.72 g C.

1.08 g of H2O contains 0.12 g H.

$\therefore$ Mass of oxygen present = 1.80 – (0.72 +0.12) = 0.96 g % of O =

{{0.96} \over {1.80}} \times 100

= 53.33 %

Q28

Using the rules for significant figures, the correct answer for the expression

{{0.02858 \times 0.112} \over {0.5702}}

will be

A 0.005613

B 0.00561

C 0.0056

D 0.006

Correct Answer

Option B

Solution

{{0.02858 \times 0.112} \over {0.5702}}

= 0.00561 Reported answer should not be more precise than least precise term (0.112 is the least precise term with three significant figures) in calculations, so there should be three significant figures in reported answer.

Q29

The ammonia (NH3) released on quantitative reaction of 0.6 g urea (NH2CONH2) with sodium hydroxide (NaOH) can be neutralized by :

A 200 ml of 0.02 N HCl

B 100 ml of 0.2 N HCl

C 100 ml of 0.1 HCl

D 200 ml of 0.4 N HCl

Correct Answer

Option B

Solution

NH2CONH2 $\to$ NH3 Using Principle of Atom Conservation 2 $\times$ moles of urea = 1 $\times$ moles of NH3 $\Rightarrow$ 2 $\times$

{{0.6} \over {60}}

$\Rightarrow$ moles of NH3 = 0.02 Also moles of NH3 = moles of HCl, because they react in 1 : 1 ratio. 100 ml of 0.2 N HCl =

{{100 \times 0.2} \over {1000}}

= 0.02 mole of HCl So option (B) is correct.

Q30

120 g of an organic compound that contains only carbon and hydrogen gives 330 g of CO2 and 270 g of water on complete combustion. The percentage of carbon and hydrogen, respectively are

A 25 and 75

B 40 and 60

C 60 and 40

D 75 and 25

Correct Answer

Option D

Solution

{C_x}{H_y} + \left( {x + {y \over 4}} \right){O_2} \to xC{O_2} + {y \over 2}{H_2}O

From the reaction, Produced CO2 = x mol and produced H2O =

{y \over 2}

mol Given produced CO2 = 330 g $\therefore$ moles of CO2

= {{330} \over {44}} = {{30} \over 4} = x

Also given produced H2O = 270 gm $\therefore$ Moles of H2O

= {{270} \over {18}} = 15 = {y \over 2}

. $\Rightarrow$ y = 30 $\therefore$

x:y = {{30} \over 4}:30 = 1:4

Formula of the compound =

{(C{H_4})_n}

$\therefore$ Weight of C in

{(C{H_4})_n}

= 12 n Weight of H in

{(C{H_4})_n}

= 4 n $\therefore$ Weight ratio of C and H = 12 n : 4 n = 3 : 1 $\therefore$ % of C =

{3 \over 4}

$\times$ 100 = 75 and % of H =

{1 \over 4}

$\times$ 100 = 25