Some Basic Concepts of Chemistry — Page 9 | NEET Chemistry

Q81

What is the mass ratio of ethylene glycol (

\mathrm{C_2H_6O_2}

, molar mass = 62 g/mol) required for making 500 g of 0.25 molal aqueous solution and 250 mL of 0.25 molar aqueous solution?

A 1 : 2

B 1 : 1

C 2 : 1

D 3 : 1

Correct Answer

Option C

Solution

For 500 g of 0.25 molal aqueous solution, 0.25 =

{{{{{w_1}} \over {62}}} \over {{{500} \over {1000}}}}

=

{{2{w_2}} \over {62}}

......(1) For 250 mL of 0.25 molar aqueous solution

{{{{{w_2}} \over {62}}} \over {{{250} \over {1000}}}} = {{4{w_2}} \over {62}}

.......(2) Dividing equation (1) by (2), we get

{{0.25} \over {0.25}} = {{{{2{w_1}} \over {62}}} \over {{{4{w_2}} \over {62}}}}

\Rightarrow {{2{w_1}} \over {62}} = {{4{w_2}} \over {62}}

\Rightarrow {{{w_1}} \over {{w_2}}} = {4 \over 2} = {2 \over 1}

Note : (1) Molality of solution is defined as number of moles of solute present per kg of solvent.

m = {{no.\,of\,moles\,(solute)} \over {weight(in\,kg) of solvent}}

(2) Molarity of solution is defined as number of moles of solute present per litre of solution.

M = {{no.\,of\,moles\,(solute)} \over {v(in\,litre\,of\,solution)}}

Q82

\mathrm{N}_{2(\mathrm{~g})}+3 \mathrm{H}_{2(\mathrm{~g})} \rightleftharpoons 2 \mathrm{NH}_{3(\mathrm{~g})}

20 \mathrm{~g} \quad ~~~5 \mathrm{~g}

Consider the above reaction, the limiting reagent of the reaction and number of moles of

\mathrm{NH}_{3}

formed respectively are :

A

\mathrm{H}_{2}, 1.42

moles

B

\mathrm{H}_{2}, 0.71

moles

C

\mathrm{N}_{2}, 1.42

moles

D

\mathrm{N}_{2}, 0.71

moles

Correct Answer

Option C

Solution

$\underset{20 \,g}{\mathrm{N}_{2(\mathrm{~g})}}+\underset{5 \mathrm{~g}}{3 H_{2(g)}} \rightleftharpoons 2 \mathrm{NH}_{3(\mathrm{~g})}$ Ideally $28 \mathrm{~g} \mathrm{~N}_{2}$ reacts with $6 \mathrm{~g} \,\mathrm{H}_{2}$ limiting reagent is $\mathrm{N}_{2}$ $\therefore$ Amount of $\mathrm{NH}_{3}$ formed on reacting $20 \mathrm{~g} \mathrm{~N}_2$ is,

\begin{aligned} &=\frac{34 \times 20}{28}=24.28 \mathrm{~g} \\\\ &=1.42 \text{ moles } \end{aligned}

Q83

Molality

(\mathrm{m})

of

3 \mathrm{M}

aqueous solution of

\mathrm{NaCl}

is : (Given : Density of solution

=1.25 \mathrm{~g} \mathrm{~mL}^{-1}

, Molar mass in

\mathrm{g} \mathrm{~mol}^{-1}: \mathrm{Na}-23, \mathrm{Cl}-35.5

)

A 2.90 m

B 3.85 m

C 1.90 m

D 2.79 m

Correct Answer

Option D

Solution

To calculate the molality $(\mathrm{m})$ of a 3M (molar) aqueous solution of $\mathrm{NaCl}$ , we need to follow certain steps, using the given data and understanding the definitions properly.

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

Molality $(\mathrm{m})$ , on the other hand, is defined as the number of moles of solute per kilogram of solvent.

The given density of the solution is $1.25 \mathrm{~g} \mathrm{~mL}^{-1}$ which allows us to calculate the mass of the solvent in a given volume of solution.

We know: Molarity (M) of NaCl = $3 \mathrm{~M}$ Density of NaCl solution = $1.25 \mathrm{~g} \mathrm{~mL}^{-1}$ Molar mass of $\mathrm{NaCl} = \mathrm{Na} + \mathrm{Cl} = 23 + 35.5 = 58.5 \mathrm{~g} \mathrm{~mol}^{-1}$ First, to find the molality, we need to find the moles of NaCl in a given mass of the solvent.

Keeping in mind that $3 \mathrm{~M}$ means there are $3 \mathrm{~moles}$ of $\mathrm{NaCl}$ per liter of solution, we consider 1 liter (or 1000 mL) of solution to simplify our calculation, knowing that density can convert volume to mass directly.

The mass of 1 liter of NaCl solution is $1.25 \mathrm{~g} \mathrm{~mL}^{-1} \times 1000 \mathrm{~mL} = 1250 \mathrm{~g}$ .

Since we have a 3 M solution of NaCl, $3 \mathrm{~moles} = \dfrac{\text{mass of the solution} - \text{mass of solvent}}{\text{molar mass of NaCl}}$ With 3 moles of NaCl, the mass of NaCl present in 1 L solution $= 3 \mathrm{~moles} \times 58.5 \mathrm{~g/mol} = 175.5 \mathrm{~g}$ The mass of the solvent (water) can be found by subtracting the mass of NaCl from the total mass of the solution.

$\text{Mass of solvent} = 1250 \mathrm{~g} - 175.5 \mathrm{~g} = 1074.5 \mathrm{~g}$ To convert this mass into kilograms (since molality is expressed per kilogram of solvent), $1074.5 \mathrm{~g} = 1.0745 \mathrm{~kg}$ Finally, the molality $(\mathrm{m})$ , $\mathrm{m} = \dfrac{\text{moles of solute}}{\text{kilograms of solvent}} = \dfrac{3 \mathrm{~moles}}{1.0745 \mathrm{~kg}}$ $\mathrm{m} \approx 2.79$ So, the correct answer is Option D: 2.79 m

Q84

25 g of an unknown hydrocarbon upon burning produces 88 g of CO2 and 9 g of H2O. This unknown hydrocarbon contains :

A 18 g of carbon and 7 g of hydrogen

B 20 g of carbon and 5 g of hydrogen

C 22 g of carbon and 3 g of hydrogen

D 24 g of carbon and 1 g of hydrogen

Correct Answer

Option D

Solution

CxHy +

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

O2 $\to$ xCO2 +

{y \over 2}

H2O No of moles of CO2 produced =

{{88} \over {44}}

= 2 mol No of moles of H2O produced =

{9 \over {18}}

= 0.5 mol According to stoichiometric mole mole analysis :

{{moles\,of\,C{O_2}} \over x} = {{moles\,of\,{H_2}O} \over {{y \over 2}}}

$\Rightarrow$

{2 \over x} = {{0.5} \over {{y \over 2}}}

$\Rightarrow$

{x \over y} = {2 \over 1}

So,

{{no\,of\,atoms\,of\,'C'\,in\,{C_x}{H_y}} \over {no\,of\,atoms\,of\,'H'\,in\,{C_x}{H_y}}} = {2 \over 1}

$\therefore$

{{Mass\,of\,'C'\,in\,{C_x}{H_y}} \over {Mass\,of\,'H'\,in\,{C_x}{H_y}}} = {{2 \times 12} \over {1 \times 1}} = {{24} \over 1}

Q85

For the following reaction, in the mass of water produced from 445 g of C57H110O6 is : 2C57H110O6(s) + 163 O2(g)

\to

114 CO2(g) + 110 H2O(l)

A 490 g

B 445 g

C 495 g

D 890 g

Correct Answer

Option C

Solution

moles of C57H110O6(s) =

{{445} \over {890}}

= 0.5 moles 2C57H110O6(s) + 163 O2(g) $\to$ 114 CO2(g) + 110 H2O(l) nH2O =

{{110} \over 4}

=

{{55} \over 2}

mH2O =

{{55} \over 2}

$\times$ 18 = 495 gm

Q86

If a rocket runs on a fuel (C15H30) and liquid oxygen, the weight of oxygen required and CO2 released for every litre of fuel respectively are : (Given : density of the fuel is 0.756 g/mL)

A 1188 g and 1296 g

B 2376 g and 2592 g

C 2592 g and 2376 g

D 3429 g and 3142 g

Correct Answer

Option C

Solution

C15H30 +

{{45} \over 2}

O2 $\to$ 15CO2 + 15H2O Given, volume of fuel = 1L = 1000 ml And density of fuel = 0.756 g/ml We know,

d = {w \over v}

\Rightarrow 0.756 = {w \over {1000}}

$\Rightarrow$ w = 756 gm $\therefore$ weight of fuel = 756 gm Molar mass of C15H30 = 15 $\times$ 12 + 30 = 210 $\therefore$ Moles of C15H30 =

{{756} \over {210}}

From equation you can see, 1 mole of C15H30 react with

{{45} \over 2}

mole of O2 $\therefore$

{{756} \over {210}}

moles of C15H30 react with

{{45} \over 2} \times {{756} \over {210}}

moles of O2 $\therefore$ Moles of O2 required =

{{45} \over 2} \times {{756} \over {210}}

$\therefore$ Mass of O2 required =

{{45} \over 2} \times {{756} \over {210}}

$\times$ 32 = 2592 g Also, From 1 mole of C15H30 15 moles of CO2 formed $\therefore$ From

{{756} \over {210}}

moles of C15H30 15 $\times$

{{756} \over {210}}

moles of CO2 formed $\therefore$ Moles of CO2 formed = 15 $\times$

{{756} \over {210}}

$\therefore$ Mass of CO2 formed = 15 $\times$

{{756} \over {210}}

$\times$ 44 = 2376 g

Q87

The ammonia evolved from the treatment of 0.30 g of an organic compound for the estimation of nitrogen was passed in 100 mL of 0.1 M sulphuric acid. The excess of acid required 20 mL of 0.5 M sodium hydroxide solution for complete neutralization. The organic compound is

A urea

B benzamide

C acetamide

D thiourea

Correct Answer

Option A

Solution

Initially total H2SO4 present = 100 mL of 0.1 M =

{{{100} \over {1000}} \times 0.1}

mole = 0.01 mole 2NaOH + H2SO4 $\to$ Na2SO4 + 2H2O Let in this reaction H2SO4 required n mole $\therefore$

{{{{20} \over {1000}} \times 0.5} \over 2}

=

{n \over 1}

\Rightarrow n

= 0.005 mole So now remaining H2SO4 = 0.01 - 0.005 = 0.005 mole .

Those remaining H2SO4 will react with NH3.

2NH3 + H2SO4 $\to$ (NH4)2SO4 Let no moles of NH3 produce through this reaction is =

x

$\therefore$

{x \over 2}

=

{{0.005} \over 1}

\Rightarrow x

= 0.01 In NH3 no of N atom is 1 and H atom is 3.

So no of moles of N atom in NH3 = 0.01 $\times$ 1 = 0.01 mole This 0.01 mole or 0.01 $\times$ 14 gm N is produced from 0.3 gm unknown organic compound. $\therefore$ % of N in unknown compound is =

{{0.01 \times 14} \over {0.3}} \times 100

= 46.6 % of N in urea [(NH4)2CO] =

{{14 \times 2} \over {60}} \times 100

= 46.6 % [ Mol weight of urea = 60] % of N in benzamide [C6H5CONH2] =

{14 \over {121}} \times 100

= 11.5 % [ Mol weight of benzamide [C6H5CONH2] = 121] % of N in acetamide [CH3CONH2] =

{14 \over {59}} \times 100

= 23.4 % [ Mol weight of acetamide [CH3CONH2] = 59] % of N in thiourea [NH2CONH2] =

{{14 \times 2} \over {76}} \times 100

= 36.8 % [ Mol weight of thiourea [NH2CSNH2] = 76] $\therefore$ compound is urea.

Q88

In Dumas' method for estimation of nitrogen 0.4 g of an organic compound gave 60 mL of nitrogen collected at 300 K temperature and 715 mm Hg pressure. The percentage composition of nitrogen in the compound is (Given : Aqueous tension at

300 \mathrm{~K}=15 \mathrm{~mm} \mathrm{Hg}

)

A

15.71 \%

B

17.46 \%

C

7.85 \%

D

20.95 \%

Correct Answer

Option A

Solution

Step 1: Calculate the pressure of dry nitrogen gas The nitrogen gas was collected over water, so we subtract the aqueous tension (pressure of water vapor) from the total pressure to find the pressure of dry nitrogen gas.

\text{Pressure of dry } \mathrm{N}_2 = 715\,\mathrm{mm\,Hg} - 15\,\mathrm{mm\,Hg} = 700\,\mathrm{mm\,Hg}

Now, convert this pressure from mm Hg to atm:

700\,\mathrm{mm\,Hg} \div 760 = \frac{700}{760}\,\mathrm{atm}

Step 2: Use the ideal gas equation to find moles of nitrogen The volume of nitrogen ( $V$ ) is given as 60 mL.

Convert this to liters: 60 mL = 0.060 L (or use $60 \times 10^{-3}$ L).

The temperature ( $T$ ) is 300 K, and the gas constant ( $R$ ) is 0.0821 L·atm·K−1·mol−1.

Using the formula $\mathrm{n} = \dfrac{\mathrm{PV}}{\mathrm{RT}}$ :

\mathrm{n} = \frac{700 \times 60 \times 10^{-3}}{760 \times 0.0821 \times 300}

When you calculate this, you get:

\mathrm{n} = 0.0022\;\text{mole}

Step 3: Calculate the mass of nitrogen gas produced Moles of nitrogen are 0.0022.

The molar mass of $\mathrm{N}_2$ is 28 g/mol.

\text{Mass of } \mathrm{N}_2 = 0.0022 \times 28 = 0.063\,\mathrm{g}

Step 4: Find the percentage of nitrogen in the compound The mass of organic compound taken is 0.4 g.

To find the percentage of nitrogen:

\text{Percentage of nitrogen} = \frac{\text{Mass of } \mathrm{N}_2}{\text{Mass of compound}} \times 100

= \frac{0.063}{0.4} \times 100 = 15.71\%

Q89

A 10 mg effervescent tablet containing sodium bicarbonate and oxalic acid releases 0.25 ml of CO2 at T = 298.15 K and p = 1 bar. If molar volume of CO2 is 25.0 L under such condition, what is the percentage of sodium bicarbonate in each tablet ? [Molar mass of NaHCO3 = 84 g mol–1]

A 33.6

B 0.84

C 8.4

D 16.8

Correct Answer

Option C

Solution

2NaHCO3 + (COOH)2 $\to$ (COONa)2 + 2H2O + 2CO2 $\therefore$ 1 mole of CO2 is produced by 1 mole of NaHCO3.

Given, volume of CO2 produced = 0.25 ml 25 L of CO2 contains 1 mol $\therefore$ 0.25 ml of CO2 contains =

{1 \over {25 \times {{10}^3}}} \times 0.25

moles = 10-5 moles $\therefore$ Moles of NaHCO3 = 10-5 moles Mass of NaHCO3 = 10-5 $\times$ 84 g $\therefore$ % of NaHCO3 in a tablet =

{{84 \times {{10}^{ - 5}}} \over {10 \times {{10}^{ - 3}}}} \times 100

= 8.4 %

Q90

The exact volumes of 1 M NaOH solution required to neutralise 50 mL of 1 M H3PO3 solution and 100 mL of 2 M H3PO2 solution, respectively, are :

A 100 mL and 50 mL

B 100 mL and 200 mL

C 100 mL and 100 mL

D 50 mL and 50 mL

Correct Answer

Option B

Solution

Millimoles of H3PO3 = M $\times$ V = 1 $\times$ 50 = 50 For 1 millimole of H3PO3, we require 2 millimoles of NaOH.

For 50 millimole of H3PO3, we require (2 $\times$ 50) = 100 millimoles of NaOH.

Millimoles of NaOH = M $\times$ V = 100 1 $\times$ V = 100 V = 100 mL Millimoles of H3PO2 = M $\times$ V = 2 $\times$ 100 = 200 For 1 millimole of H3PO2, we require 1 millimoles of NaOH.

For 200 millimoles of H3PO2, we require 200 millimoles of NaOH.

So, volume of NaOH = 200 mL