Some Basic Concepts of Chemistry — Page 8 | JEE Chemistry

Q71

The number of molecules and moles in 2.8375 litres of O

_2

at STP are respectively

A 1.505

\times

10

^{23}

and 0.250 mol

B 7.527

\times

10

^{22}

and 0.250 mol

C 7.527

\times

10

^{23}

and 0.125 mol

D 7.527

\times

10

^{22}

and 0.125 mol

Correct Answer

Option D

Solution

At STP, one mole of any gas occupies 22.4 liters.

Therefore, 2.8375 liters of oxygen gas at STP is equal to 0.125 moles.

The number of molecules of oxygen gas in 0.125 moles is calculated by multiplying the number of moles by Avogadro's number, which is

6.022 × 10^{23}

molecules/mol. This gives us a total of

7.527 × 10^{22}

molecules of oxygen gas in 2.8375 liters at STP. Here is the calculation:

\begin{align} &\text{Number of moles of }O_2 = \frac{2.8375 \text{ L}}{22.4 \text{ L/mol}} = 0.125 \text{ mol} \\\\ &\text{Number of molecules of }O_2 = 0.125 \text{ mol} \times 6.022 \times 10^{23} \text{ molecules/mol} = 7.527 \times 10^{22} \text{ molecules} \end{align}

Q72

The average molar mass of chlorine is 35.5 g mol–1. The ratio of 35Cl to 37Cl in naturally occuring chlorine is close to :

A 1 : 1

B 2 : 1

C 3 : 1

D 4 : 1

Correct Answer

Option C

Solution

Average molar mass =

{{{n_1}{M_1} + {n_2}{M_2}} \over {\left( {{n_1} + {n_2}} \right)}}

$\therefore$ 35.5 =

{{35x + 37y} \over {x + y}}

$\Rightarrow$ 1.5y = –0.5x $\Rightarrow$

{x \over y} = {3 \over 1}

Q73

A sample of

\mathrm{CaCO}_3

and

\mathrm{MgCO}_3

weighed

2.21 \mathrm{~g}

is ignited to constant weight of

1.152 \mathrm{~g}

. The composition of mixture is : (Given molar mass in

\mathrm{g} \mathrm{~mol}^{-1} \mathrm{CaCO}_3: 100, \mathrm{MgCO}_3: 84

)

A

1.187 \mathrm{~g} \mathrm{~CaCO}_3+1.187 \mathrm{~g} \mathrm{~MgCO}_3

B

1.187 \mathrm{~g} \mathrm{~CaCO}_3+1.023 \mathrm{~g} \mathrm{~MgCO}_3

C

1.023 \mathrm{~g} \mathrm{~CaCO}_3+1.187 \mathrm{~g} \mathrm{~MgCO}_3

D

1.023 \mathrm{~g} \mathrm{~CaCO}_3+1.023 \mathrm{~g} \mathrm{~MgCO}_3

Correct Answer

Option B

Solution

\begin{aligned} & \mathrm{CaCO}_3(\mathrm{s}) \xrightarrow{\Delta} \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_2(\mathrm{g}) \\ & \mathrm{MgCO}_3(\mathrm{s}) \xrightarrow{\Delta} \mathrm{MgO}(\mathrm{s})+\mathrm{CO}_2(\mathrm{g}) \end{aligned}

Let the weight of

\mathrm{CaCO}_3

be

\mathrm{x}

gm $\therefore$ weight of

\mathrm{MgCO}_3=(2.21-\mathrm{x}) \mathrm{~gm}

Moles of

\mathrm{CaCO}_3

decomposed $=$ moles of

\mathrm{CaO}

formed

\frac{\mathrm{x}}{100}=

moles of

\mathrm{CaO}

formed $\therefore$ weight of

\mathrm{CaO}

formed

=\frac{\mathrm{x}}{100} \times 56

Moles of

\mathrm{MgCO}_3

decomposed $=$ moles of

\mathrm{MgO}

formed

\frac{(2.21-x)}{84}=

moles of

\mathrm{MgO}

formed $\therefore$ weight of

\mathrm{MgO}

formed

=\frac{2.21-\mathrm{x}}{84} \times 40

\Rightarrow \frac{2.21-\mathrm{x}}{84} \times 40+\frac{\mathrm{x}}{100} \times 56=1.152

\therefore \mathrm{x}=1.1886 \mathrm{~g}=

weight of

\mathrm{CaCO}_3

& weight of

\mathrm{MgCO}_3=1.0214 \mathrm{~g}

Q74

Given below are two statements: one is labelled as Assertion

\mathbf{A}

and the other is labelled as Reason

\mathbf{R}

Assertion A :

3.1500 \mathrm{~g}

of hydrated oxalic acid dissolved in water to make

250.0 \mathrm{~mL}

solution will result in

0.1 \mathrm{~M}

oxalic acid solution. Reason

\mathbf{R}

: Molar mass of hydrated oxalic acid is

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

In the light of the above statements, choose the correct answer from the options given below.

A Both A and R are true but R is NOT the correct explanation of A

B A is true but R is false

C Both A and R are true and R is the correct explanation of A

D A is false but R is true

Correct Answer

Option C

Solution

Firstly, let's calculate the molarity of the oxalic acid solution given in the Assertion (A).

The molarity (M) is defined as the number of moles of solute per liter of solution.

The number of moles of solute is given by the mass of the solute divided by its molar mass.

So, the number of moles of hydrated oxalic acid is

\frac{3.1500 \, \text{g}}{126 \, \text{g/mol}} = 0.025 \, \text{mol}

And the volume of the solution is 250.0 mL, or 0.250 L. Therefore, the molarity of the solution is

\frac{0.025 \, \text{mol}}{0.250 \, \text{L}} = 0.1 \, \text{M}

This matches the molarity given in the Assertion (A), so the Assertion is true.

The Reason (R) gives the molar mass of hydrated oxalic acid as 126 g/mol, which was used in the calculation above.

Therefore, the Reason is also true.

Moreover, the Reason (R) is the correct explanation of Assertion (A), as it provides the necessary information (the molar mass of hydrated oxalic acid) to calculate the molarity of the solution.

So, the correct option is C: Both A and R are true and R is the correct explanation of A.

Q75

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of the substance will

A be a function of the molecular mass of the substance

B remain unchanged

C increase two fold

D decrease twice

Correct Answer

Option D

Solution

Let Relative Atomic Mass is in short R.A.M then R.A.M =

{{mass\,of\,1\,atom} \over {{1 \over {12}} \times mass\,of\,1\,atom\,of\,{C^{12}}}}

$\Rightarrow$

{{mass\,of\,1\,atom} \over {mass\,of\,1\,atom\,of\,{C^{12}}}}

$\times$ 12 Now if we take 1/6 instead of 1/12 at first then the equation will become $\Rightarrow$

{{mass\,of\,1\,atom} \over {mass\,of\,1\,atom\,of\,{C^{12}}}}

$\times$ 6 as 6 is half of 12 then R.A.M will decrease twice.

Q76

Hemoglobin contains

0.34 \%

of iron by mass. The number of Fe atoms in

3.3 \mathrm{~g}

of hemoglobin is (Given: Atomic mass of Fe is

56 \,\mathrm{u}, \mathrm{N}_{\mathrm{A}}=6.022 \times 10^{23} \mathrm{~mol}^{-1}

.)

A

1.21 \times 10^{5}

B

12.0 \times 10^{16}

C

1.21 \times 10^{20}

D

3.4 \times 10^{22}

Correct Answer

Option C

Solution

According to the question,

100 \mathrm{~g}

of hemoglobin contains

0.34 \mathrm{~g}

of iron

3.3 \mathrm{~g}

of hemoglobin contains

\frac{0.34}{100} \times 3.3 \mathrm{~g}

of iron moles of

\mathrm{Fe}=\frac{0.34 \times 3.3}{100 \times 56}=\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{A}}}

$N=\dfrac{0.34 \times 3.3 \times 6.022 \times 10^{23}}{100 \times 56}$

=1.21 \times 10^{20}

Q77

Consider the reaction

4 \mathrm{HNO}_{3}(1)+3 \mathrm{KCl}(\mathrm{s}) \rightarrow \mathrm{Cl}_{2}(\mathrm{~g})+\mathrm{NOCl}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})+3 \mathrm{KNO}_{3}(\mathrm{~s})

The amount of

\mathrm{HNO}_{3}

required to produce

110.0 \mathrm{~g}

of

\mathrm{KNO}_{3}

is (Given: Atomic masses of

\mathrm{H}, \mathrm{O}, \mathrm{N}

and

\mathrm{K}

are

1,16,14

and 39, respectively.)

A 32.2 g

B 69.4 g

C 91.5 g

D 162.5 g

Correct Answer

Option C

Solution

\begin{array}{r} 4 \mathrm{HNO}_{3}(\mathrm{l})+3 \mathrm{KCl}(\mathrm{s}) \longrightarrow \mathrm{Cl}_{2}(\mathrm{~g})+\mathrm{NOCl}(\mathrm{g})+ \\ 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})+3 \mathrm{KNO}_{3}(\mathrm{~s}) \end{array}

$\because 110 \mathrm{~g}$ of $\mathrm{KNO}_{3} \Rightarrow$ moles of $\mathrm{KNO}_{3}=\dfrac{110}{101}$

=1.089 \mathrm{~mol}

As, 4 mole of $\mathrm{HNO}_{3}$ produces $3 \mathrm{~mol}$ of $\mathrm{KNO}_{3}$ .

Hence, the moles of $\mathrm{HNO}_{3}$ required to produce $1.089$ moles of $\mathrm{KNO}_{3}=\dfrac{4}{3} \times 1.089=1.452 \mathrm{~mol}$ Hence, mass of $\mathrm{HNO}_{3}$ required is $1.452 \times 63$

=91.5 \mathrm{~g} \text{ }

Q78

Match List I with List II .tg .tg List - I List - II (

\Delta_0

) A. 16 g of

\mathrm{CH_4~(g)}

I. Weighs 28 g B. 1 g of

\mathrm{H_2~(g)}

II.

60.2\times10^{23}

electrons C. 1 mole of

\mathrm{N_2~(g)}

III. Weighs 32 g D. 0.5 mol of

\mathrm{SO_2~(g)}

IV. Occupies 11.4 L volume of STP Choose the correct answer from the options given below:

A A-II, B-III, C-IV, D-I

B A-II, B-IV, C-I, D-III

C A-I, B-III, C-II, D-IV

D A-II, B-IV, C-III, D-I

Correct Answer

Option B

Solution

A. 16 g of $\mathrm{CH_4~(g)}$ The molar mass of $\mathrm{CH_4}$ (Methane) is 16 g/mol.

Therefore, 16 g of $\mathrm{CH_4}$ is equivalent to 1 mole of $\mathrm{CH_4}$ .

Furthermore, each molecule of $\mathrm{CH_4}$ has 10 electrons (6 from Carbon and 4 from Hydrogen).

As a result, 1 mole of $\mathrm{CH_4}$ (or $6.022 \times 10^{23}$ molecules of $\mathrm{CH_4}$ ) would have $10 \times 6.022 \times 10^{23} = 60.2 \times 10^{23}$ electrons.

This matches with (II).

B. 1 g of $\mathrm{H_2~(g)}$ The molar mass of $\mathrm{H_2}$ (Hydrogen) is 2 g/mol.

Therefore, 1 g of $\mathrm{H_2}$ is equivalent to 0.5 moles of $\mathrm{H_2}$ .

The volume that a given quantity of gas occupies is proportional to the number of moles of gas.

At standard temperature and pressure (STP), 1 mole of any ideal gas occupies 22.4 liters.

Therefore, 0.5 moles of gas would occupy $0.5 \times 22.4 = 11.2$ liters.

The closest match is (IV) with 11.4 liters (the small discrepancy may be due to rounding or slightly different conditions than STP).

C. 1 mole of $\mathrm{N_2~(g)}$ The molar mass of $\mathrm{N_2}$ (Nitrogen) is 28 g/mol.

Therefore, 1 mole of $\mathrm{N_2}$ weighs 28 g.

This matches with (I).

D.

0.5 mol of $\mathrm{SO_2~(g)}$ The molar mass of $\mathrm{SO_2}$ (Sulfur Dioxide) is 64 g/mol.

Therefore, 0.5 moles of $\mathrm{SO_2}$ weigh $0.5 \times 64 = 32$ g.

This matches with (III).

Thus, the correct matches are: A - II B - IV C - I D - III

Q79

A gaseous hydrocarbon gives upon combustion 0.72 g of water and 3.08 g of CO2. The empirical formula of the hydrocarbon is

A C2H4

B C3H4

C C6H5

D C7H8

Correct Answer

Option D

Solution

Required reaction, C

x

H

y

+

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

O2 $\to$

x

CO2 +

{{y \over 2}}

H2O 0.72 gm of H2O =

{{{0.72} \over {18}}}

mole of H2O = 0.04 mole of H2O In one H2O molecule 2 hydrogen atoms present.

So in 0.04 mole of H2O molecules 2 $\times$ 0.04 = 0.08 moles of H atoms present.

0.72 gm of CO2 =

{{{3.08} \over {44}}}

mole of CO2 = 0.07 mole of CO2 And in one CO2 molecule 1 C atom present.

So in 0.07 mole of CO2 molecules 0.07 $\times$ 1 = 0.07 moles of C atoms present $\therefore$ C : H = 0.07 : 0.08 = 7 : 8 $\therefore$ Empirical formula of hydrocarbon = C7H8

Q80

Which of the following have same number of significant figures? A. 0.00253 B. 1.0003 C. 15.0 D. 163 Choose the correct answer from the options given below

A A, C and D only

B B and C only

C A, B and C only

D C and D only

Correct Answer

Option A

Solution

Significant figures are the digits in a number that carry meaningful information about its precision.

In other words, they are the numbers that are known with certainty.

Here is how we calculate significant figures for each number: A.

0.00253: There are 3 significant figures here (253).

Leading zeroes do not count as significant figures.

B.

1.0003: There are 5 significant figures here.

All non-zero digits are significant, and zeroes between non-zero digits or at the end of the number and after the decimal point are significant.

C.

15.0: There are 3 significant figures here.

The zero after the decimal point counts as it indicates the precision of the measurement.

D.

163: There are 3 significant figures here.

All non-zero digits are significant.

So, comparing these, we see that options A (0.00253), C (15.0), and D (163) each have 3 significant figures.