Some Basic Concepts of Chemistry — Page 8 | NEET Chemistry

Q71

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}

Q72

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}

Q73

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.

Q74

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.

Q75

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}

Q76

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{ }

Q77

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

Q78

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

Q79

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.

Q80

Choose the correct statements.(A) Weight of a substance is the amount of matter present in it.(B) Mass is the force exerted by gravity on an object.(C) Volume is the amount of space occupied by a substance.(D) Temperatures below 0°C are possible in Celsius scale, but in Kelvin scale negative temperature is not possible.(E) Precision refers to the closeness of various measurements for the same quantity.Choose the correct answer from the options given below :

A (B), (C) and (D) Only

B (A), (B) and (C) Only

C (C), (D) and (E) Only

D (A), (D) and (E) Only

Correct Answer

Option C

Solution

A) Weight of a substance is the amount of matter present in it.

Not correct.

The amount of matter present in a substance is called its mass, not its weight.

Mass of a substance is the amount of matter present in it.

B) Mass is the force exerted by gravity on an object.

Not correct.

Mass is not the force exerted on an object by gravity.

Mass is the amount of matter in an object, while weight is the force of gravity on that object.

C) Volume is the amount of space occupied by a substance.

Correct.

The amount of space occupied by a substance is its volume, which is simply the equivalent three dimensional measurements of spatial geometry (height, width; and (length).

Standard units of volume are cubic centimeters for solids, milliliters or liters for liquids, and liters pr gases.

D) Temperatures below $0^{\circ} \mathrm{C}$ are possible in Celsius scale, but in Kelvin scale negative temperature is not possible.

Correct.

Celsius temp is possible below $0^{\circ} \mathrm{C}$ (negative temp).

But, it is not possible in the Kelvin Scale $273 \mathrm{~K} \rightarrow$ Absolute temperature Kelvin temp $={ }^{\circ} \mathrm{C}$ temp +273.15 The minimum temperature possible in the Kelvin scale is the absolute zero temperature ( $\mathrm{OK}$ ).

OK means $-273.15^{\circ} \mathrm{C}$ .

Since absolute zero is the coldest possible temperature, there we no negative values on the Kelvin temperature scale.

E) Precision refers to the closeness of various measurements for the same quantity.

Correct.

A measure of how close repeated measurements are to each other is known as precision.

Precision is the reproducibility of a result or measurement.

If weight of a substance is measured five times and get 8.2 kg each time, then the measurement is very precise - Example.

So, the correct statements are (C) , (D) and (E).

Correct answer: Option 3) (C) ,(D) and(E) only