Some Basic Concepts of Chemistry — Page 6 | JEE Chemistry

Q51

The number of moles of methane required to produce

11 \mathrm{~g} \mathrm{~CO}_2(\mathrm{g})

after complete combustion is : (Given molar mass of methane in

\mathrm{g} \mathrm{~mol}^{-1}: 16

)

A 0.35

B 0.25

C 0.5

D 0.75

Correct Answer

Option B

Solution

To find the number of moles of methane ( $CH_4$ ) required to produce 11g of $CO_2$ upon complete combustion, we first need the balanced chemical equation for the combustion of methane:

CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O

From the balanced equation, we see that 1 mole of $CH_4$ produces 1 mole of $CO_2$ .

Therefore, the moles of $CH_4$ needed to produce a certain amount of $CO_2$ will be equal to the moles of $CO_2$ produced.

To find the moles of $CO_2$ produced from 11g of $CO_2$ , we use the formula:

\text{Moles} = \frac{\text{Mass}}{\text{Molar mass}}

The molar mass of $CO_2$ is $44 \mathrm{~g/mol}$ (12 from carbon and 16 $\times$ 2 from oxygen).

Then, the moles of $CO_2$ produced from 11g of $CO_2$ can be calculated as:

\text{Moles of } CO_2 = \frac{11 \mathrm{~g}}{44 \mathrm{g/mol}} = 0.25 \mathrm{~mol}

Since the stoichiometry of the reaction between methane and oxygen is 1:1 for $CH_4$ and $CO_2$ , the moles of $CH_4$ required to produce 11g of $CO_2$ is also 0.25 moles.

Therefore, the correct answer is Option B: 0.25.

Q52

On complete combustion 1.0 g of an organic compound

(\mathrm{X})

gave 1.46 g of

\mathrm{CO}_2

and 0.567 g of

\mathrm{H}_2 \mathrm{O}

. The empirical formula mass of compound

(\mathrm{X})

is __________ g. (Given molar mass in

\mathrm{g} \mathrm{mol}^{-1} \mathrm{C}: 12, \mathrm{H}: 1, \mathrm{O}: 16

)

A 60

B 45

C 30

D 15

Correct Answer

Option C

Solution

Calculate the Moles of Carbon: Carbon is fully converted to $\mathrm{CO}_2$ .

Moles of $\mathrm{CO}_2 = \dfrac{1.46\, \text{g}}{44\, \text{g/mol}} = 0.033 \text{ mol}$ .

Moles of Carbon ( $\text{C}$ ) = Moles of $\mathrm{CO}_2$ = 0.033 mol.

Mass of Carbon = $0.033 \times 12\, \text{g/mol} = 0.396\, \text{g}$ .

Calculate the Moles of Hydrogen: Hydrogen is fully converted to $\mathrm{H}_2\mathrm{O}$ .

Moles of $\mathrm{H}_2\mathrm{O} = \dfrac{0.567\, \text{g}}{18\, \text{g/mol}} = 0.0315 \text{ mol}$ .

Moles of Hydrogen ( $\text{H}$ ) = $2 \times 0.0315 = 0.063 \text{ mol}$ .

Mass of Hydrogen = $0.063 \times 1\, \text{g/mol} = 0.063\, \text{g}$ .

Calculate the Mass of Oxygen: Total mass of compound $\mathrm{X}$ = 1.0 g.

Mass of Oxygen = Total mass - (Mass of Carbon + Mass of Hydrogen) Mass of Oxygen = $1.0\, \text{g} - (0.396\, \text{g} + 0.063\, \text{g}) = 0.541\, \text{g}$ .

Moles of Oxygen ( $\text{O}$ ) = $\dfrac{0.541}{16} = 0.0338 \approx 0.033 \text{ mol}$ .

Determine the Empirical Formula: The ratio of moles: $\text{C} = 0.033$ , $\text{H} = 0.063$ , $\text{O} = 0.033$ .

Simplified, this ratio is approximately 1:2:1.

Empirical formula: $\mathrm{CH}_2\mathrm{O}$ .

Calculate the Empirical Formula Mass: Empirical formula mass = $12 \times 1 + 1 \times 2 + 16 \times 1 = 30\, \text{g/mol}$ .

Therefore, the empirical formula mass of compound $\mathrm{X}$ is 30 g/mol.

Q53

The density of a solution prepared by dissolving 120 g of urea (mol. Mass = 60 u ) in 1000g of water is 1.15 g/mL. The molarity of this solution is :

A 0.50 M

B 1.78 M

C 1.02 M

D 2.05 M

Correct Answer

Option D

Solution

We know molarity (M) =

{{no\,of\,moles\,of\,solute} \over {volume\,of\,solution\,in\,litre}}

Moles of solute =

{{120} \over {60}}

= 2 Mass of solution = 1000 + 120 = 1120 gm Density of solution = 1.15 g/mL $\therefore$ volume of solution =

{{1120} \over {1.15}}

= 973.9 mL =

{{973.9} \over {1000}}

litre = 0.9739 litre $\therefore$ M =

{2 \over {0.9739}}

= 2.05

Q54

With increase of temperature, which of these changes?

A molality

B weight fraction of solute

C molarity

D mole fraction

Correct Answer

Option C

Solution

Molarity = Number of moles of solute Volume of solution (in L) Molality = Number of moles of solute Mass of solvent (in kg) Mole fraction = Number of moles of component Total number of moles of all components Weight fraction of solute = Weight of a component Total weight of solution $\times$ 100 KEY Concept : Those units which are volume related will be affected by the change in temperature.

The definition of molarity is - Number of moles of solute present in one litre of solution.

From definition you can see it is depends on volume which increases with increasing temperature and decreases with decreasing temperature.

All the other three options (Molality, Mole fraction, Weight fraction of solute) are mass related units and temperature has no effect on mass.

Q55

What volume of hydrogen gas at 273 K and 1 atm pressure will be consumed in obtaining 21.6 g of elemental boron (atomic mass = 10.8) from the reduction of boron trichloride by hydrogen?

A 67.2 L

B 44.8 L

C 22.4 L

D 89.6 L

Correct Answer

Option A

Solution

The reaction is 2BCl3 + 3H2 $\to$ 2B + 6HCL $\therefore$ 3 moles of hydrogen produces 2 moles of boron So 2 $\times$ 10.8 gm = 21.6 gm of boron is produces from 3 $\times$ 22.4 = 67.2 L hydrogen

Q56

On combustion 0.210 g of an organic compound containing C, H and O gave 0.127 g H2O and 0.307 g CO2. The percentages of hydrogen and oxygen in the given organic compound respectively are:

A 7.55, 43.85

B 6.72, 53.41

C 6.72, 39.87

D 53.41, 39.6

Correct Answer

Option B

Solution

In the combustion of organic compound, all "C" in $\mathrm{CO}_2$ and all " H " in $\mathrm{H}_2 \mathrm{O}$ comes from organic compound Weight of " C " in $\mathrm{CO}_2=\dfrac{12}{44} \times 0.307$

=0.0837 \mathrm{~gm}

Weight of " H " in $\mathrm{H}_2 \mathrm{O}=\dfrac{2}{18} \times 0.127=0.0141 \mathrm{~g}$

\begin{aligned} &\begin{gathered} \% ' H^{\prime} \text{ in compound }=\frac{0.0141}{0.21} \times 100=6.719 \% \\ =6.72 \% \end{gathered}\\ &\text{ Weight of "O" in compound }\\ &\begin{aligned} & =0.210-(0.0837+0.0141) \\ & =0.1122 \end{aligned}\\ &\begin{aligned} & \% \text{ of "O" in compound }=\frac{0.1122}{0.21} \times 100 \\ & =53.41 \% \end{aligned} \end{aligned}

Q57

\mathrm{SO}_{2} \mathrm{Cl}_{2}

on reaction with excess of water results into acidic mixture

\mathrm{SO}_{2} \mathrm{Cl}_{2}+2 \mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{H}_{2} \mathrm{SO}_{4}+2 \mathrm{HCl}

16 moles of

\mathrm{NaOH}

is required for the complete neutralisation of the resultant acidic mixture. The number of moles of

\mathrm{SO}_{2} \mathrm{Cl}_{2}

used is :

A 16

B 8

C 4

D 2

Correct Answer

Option C

Solution

\mathrm{SO}_{2} \mathrm{Cl}_{2}+2 \mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{H}_{2} \mathrm{SO}_{4}+2 \mathrm{HCl}

Moles of

\mathrm{NaOH}

required for complete neutralization of resultant acidic mixture

=16

moles And 1 mole of

\mathrm{SO}_{2} \mathrm{Cl}_{2}

produced 4 moles of

\mathrm{H}^{+}

. $\therefore$ Moles of

\mathrm{SO}_{2} \mathrm{Cl}_{2}

used will be

=\frac{16}{4}=4

moles

Q58

\mathrm{CaCO}_3(\mathrm{~s})+2 \mathrm{HCl}(\mathrm{aq}) \rightarrow \mathrm{CaCl}_2(\mathrm{aq})+\mathrm{CO}_2(\mathrm{~g})+\mathrm{H}_2 \mathrm{O}(\mathrm{l})

Consider the above reaction, what mass of

\mathrm{CaCl}_2

will be formed if 250 mL of 0.76 M HCl reacts with 1000 g of

\mathrm{CaCO}_3

? (Given : Molar mass of

\mathrm{Ca}, \mathrm{C}, \mathrm{O}, \mathrm{H}

and Cl are

40,12,16,1

and

35.5 \mathrm{~g} \mathrm{~mol}^{-1}

, respectively)

A 3.908 g

B 2.636 g

C 10.545 g

D 5.272 g

Correct Answer

Option C

Solution

The reaction is: $\mathrm{CaCO}_3(\mathrm{s}) + 2\mathrm{HCl}(\mathrm{aq}) \rightarrow \mathrm{CaCl}_2(\mathrm{aq}) + \mathrm{CO}_2(\mathrm{g}) + \mathrm{H}_2\mathrm{O}(\mathrm{l})$ Calculate Moles of $\mathrm{CaCO}_3$ : The molar mass of $\mathrm{CaCO}_3$ is calculated as $40 + 12 + 3 \times 16 = 100 \ \mathrm{g/mol}$ Given 1000 g of $\mathrm{CaCO}_3$ , the number of moles is $\dfrac{1000}{100} = 10 \ \mathrm{moles}$ Calculate Moles of $\mathrm{HCl}$ : Using the given concentration and volume for $\mathrm{HCl}$ , the moles are calculated as follows: $0.76 \times \dfrac{250}{1000} = 0.19 \ \mathrm{moles}$ $\mathrm{HCl}$ is the limiting reagent (L.R.) because it has fewer moles than needed to completely react with the $\mathrm{CaCO}_3$ .

Calculate Moles of $\mathrm{CaCl}_2$ Formed: According to the stoichiometry of the reaction, 2 moles of $\mathrm{HCl}$ produce 1 mole of $\mathrm{CaCl}_2$ .

Therefore, the moles of $\mathrm{CaCl}_2$ formed are: $\dfrac{0.19}{2} = 0.095 \ \mathrm{moles}$ Calculate Mass of $\mathrm{CaCl}_2$ : The molar mass of $\mathrm{CaCl}_2$ is: $40 + 2 \times 35.5 = 111 \ \mathrm{g/mol}$ Thus, the mass of $\mathrm{CaCl}_2$ is: $0.095 \times 111 = 10.545 \ \mathrm{g}$ Hence, the mass of $\mathrm{CaCl}_2$ formed is 10.545 g.

Q59

The molarity of a solution obtained by mixing 750 mL of 0.5 (M) HCl with 250 mL of 2(M) HCl will be:

A 1.00 M

B 1. 75 M

C 0.975 M

D 0.875 M

Correct Answer

Option D

Solution

The formula for molarity of mixture of two substance is =

{{{M_1}{V_1} + {M_2}{V_2}} \over {{V_1} + {V_2}}}

Here

{M_1} = 0.5

,

{V_1} = 750

,

{M_2} = 2

,

{V_2} = 250

$\therefore$ Molarity of mixture =

{{0.5 \times 750 + 2 \times 250} \over {750 + 250}}

=

{{375 + 500} \over {1000}}

=

{{875} \over {1000}}

= 0.875

Q60

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

Correct Answer

Option C

Solution

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.