Some Basic Concepts of Chemistry Questions — NEET Chemistry

Q1

Thermal decomposition of a Mn compound (X) at 513 K results in compound Y, MnO2 and gaseous product. MnO2 reacts with NaCl and concentrated H2O4 to give a pungent gas Z. X, Y and Z, respectively, are :

A KMnO4, K2MnO4 and Cl2

B K2MnO4, KMnO4 and SO2

C K3MnO4, K2MnO4 and Cl2

D K2MnO4, KMnO4 and Cl2

Correct Answer

Option A

Solution

KMnO4

\overset{{513\,\,K}}\longrightarrow

K2MnO4 + MnO2 + O2 (X) (Y) .tg .tg MnO2 + NaCl + conc H2SO4 $\to$ MnSO4 + NaHSO4 + H2O + Cl2 (Z)

Q2

The number of hydrogen atoms present in 5.4 g of urea is: (Given: Molar mass of urea :

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

\mathrm{N}_{\mathrm{A}}: 6.022 \times 10^{23}

particles

\mathrm{mol}^{-1}

)

A

1.084 \times 10^{23}

B

1.084 \times 10^{22}

C

2.168 \times 10^{22}

D

2.168 \times 10^{23}

Correct Answer

Option D

Solution

\begin{aligned} & \text{ Mole of urea }=\frac{5.4}{60}=0.09 \\ & \begin{aligned} \text{ Number of hydrogen atoms } & =0.09 \times 4 \times 6.022 \times 10^{23} \\ & =2.1679 \times 10^{23} \\ & \simeq 2.168 \times 10^{23} \end{aligned} \end{aligned}

Q3

When

1 \mathrm{dm}^3

of CO 2 gas is passed over hot coke the volume of gaseous mixture after complete reaction at STP becomes

1.4 \mathrm{dm}^3

. The composition of the gaseous mixture at STP is :

A

0.8 \mathrm{dm}^3

of

\mathrm{CO}, 0.8 \mathrm{dm}^3

of

\mathrm{CO}_2

B

0.8 \mathrm{dm}^3

of

\mathrm{CO}, 0.6 \mathrm{dm}^3

of

\mathrm{CO}_2

C

0.6 \mathrm{dm}^3

of

\mathrm{CO}, 0.8 \mathrm{dm}^3

of

\mathrm{CO}_2

D

0.6 \mathrm{dm}^3

of

\mathrm{CO}, 0.4 \mathrm{dm}^3

of

\mathrm{CO}_2

Correct Answer

Option B

Solution

The reaction of carbon dioxide with hot coke is:

\mathrm{CO_2 + C \rightarrow 2CO}

At the same temperature and pressure, volumes of gases are in the same ratio as the number of moles.

Let $x \,\mathrm{dm^3}$ of $\mathrm{CO_2}$ react.

Then according to the equation: $x \,\mathrm{dm^3}$ of $\mathrm{CO_2}$ is consumed $2x \,\mathrm{dm^3}$ of $\mathrm{CO}$ is formed Initially, volume of $\mathrm{CO_2} = 1 \,\mathrm{dm^3}$ So after reaction: Unreacted $\mathrm{CO_2} = (1 - x)\,\mathrm{dm^3}$ Formed $\mathrm{CO} = 2x\,\mathrm{dm^3}$ Total final volume is given as $1.4 \,\mathrm{dm^3}$ So,

(1-x) + 2x = 1.4

1 + x = 1.4

x = 0.4

Now,

\text{Volume of CO} = 2x = 0.8 \,\mathrm{dm^3}

\text{Volume of CO_2} = 1 - x = 0.6 \,\mathrm{dm^3}

Hence, the gaseous mixture contains:

0.8 \,\mathrm{dm^3}\ \mathrm{of\ CO} \quad \text{and} \quad 0.6 \,\mathrm{dm^3}\ \mathrm{of\ CO_2}

So the correct option is Option B .

Q4

Dalton's Atomic theory could not explain which of the following?

A Law of multiple proportion

B Law of gaseous volume

C Law of conservation of mass

D Law of constant proportion

Correct Answer

Option B

Solution

Dalton's theory could explain the laws of chemical combination.

However, it could not explain the laws of gaseous volumes.

Q5

The amount of glucose required to prepare

250 \mathrm{~mL}

of

\dfrac{\mathrm{M}}{20}

aqueous solution is : (Molar mass of glucose :

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

)

A 2.25 g

B 4.5 g

C 0.44 g

D 1.125 g

Correct Answer

Option A

Solution

To find the amount of glucose required to prepare

250 \, \mathrm{mL}

of a

\frac{M}{20}

aqueous solution, we need to follow these steps: 1.

Convert the volume from milliliters to liters, because molarity is expressed in moles per liter (M).

2.

Use the equation for molarity:

M = \frac{n}{V}

where:

M

is the molarity (given as

\frac{1}{20} \, \mathrm{M}

)

n

is the number of moles of solute

V

is the volume of the solution in liters 3. Rearrange the equation to solve for the number of moles of glucose (

n

):

n = M \times V

4.

Convert the moles of glucose to grams using the molar mass of glucose.

Let's proceed with the calculations: 1.

Convert the volume to liters:

250 \, \mathrm{mL} = \frac{250}{1000} \, \mathrm{L} = 0.25 \, \mathrm{L}

2. Calculate the moles of glucose:

n = \left( \frac{1}{20} \right) \times 0.25 \, \mathrm{L} = \frac{0.25}{20} \, \mathrm{mol} = 0.0125 \, \mathrm{mol}

3. Convert the moles of glucose to grams using the molar mass of glucose:

\text{mass} = n \times \text{molar mass}

\text{mass} = 0.0125 \, \mathrm{mol} \times 180 \, \mathrm{g} \, \mathrm{mol}^{-1} = 2.25 \, \mathrm{g}

So, the amount of glucose required is: Option A: 2.25 g

Q6

1.0 \mathrm{~g}

of

\mathrm{H}_2

has same number of molecules as in:

A

14 \mathrm{~g}

of

\mathrm{N}_2

B

18 \mathrm{~g}

of

\mathrm{H}_2 \mathrm{O}

C

16 \mathrm{~g}

of

\mathrm{CO}

D

28 \mathrm{~g}

of

\mathrm{N}_2

Correct Answer

Option A

Solution

To determine which of the given options contains the same number of molecules as

1.0 \mathrm{~g}

of

\mathrm{H}_2

, we need to use the concept of moles and Avogadro's number. First, let's calculate the number of moles in

1.0 \mathrm{~g}

of

\mathrm{H}_2

. The molar mass of

\mathrm{H}_2

is

2 \mathrm{~g/mol}

. Therefore, the number of moles of

\mathrm{H}_2

in

1.0 \mathrm{~g}

can be calculated as:

\text{Number of moles of } \mathrm{H}_2 = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{1.0 \mathrm{~g}}{2 \mathrm{~g/mol}} = 0.5 \mathrm{~mol}

Next, we need to find out which of the given options corresponds to the same number of moles.

Let's calculate the number of moles for each option: Option A:

14 \mathrm{~g} \text{ of } \mathrm{N}_2

. The molar mass of

\mathrm{N}_2

is

28 \mathrm{~g/mol}

. The number of moles of

\mathrm{N}_2

in

14 \mathrm{~g}

is:

\text{Number of moles of } \mathrm{N}_2 = \frac{14 \mathrm{~g}}{28 \mathrm{~g/mol}} = 0.5 \mathrm{~mol}

Option B:

18 \mathrm{~g} \text{ of } \mathrm{H}_2 \mathrm{O}

. The molar mass of

\mathrm{H}_2 \mathrm{O}

is

18 \mathrm{~g/mol}

. The number of moles of

\mathrm{H}_2 \mathrm{O}

in

18 \mathrm{~g}

is:

\text{Number of moles of } \mathrm{H}_2 \mathrm{O} = \frac{18 \mathrm{~g}}{18 \mathrm{~g/mol}} = 1 \mathrm{~mol}

Option C:

16 \mathrm{~g} \text{ of } \mathrm{CO}

. The molar mass of

\mathrm{CO}

is

28 \mathrm{~g/mol}

. The number of moles of

\mathrm{CO}

in

16 \mathrm{~g}

is:

\text{Number of moles of } \mathrm{CO} = \frac{16 \mathrm{~g}}{28 \mathrm{~g/mol}} = 0.571 \mathrm{~mol}

Option D:

28 \mathrm{~g} \text{ of } \mathrm{N}_2

. The molar mass of

\mathrm{N}_2

is

28 \mathrm{~g/mol}

. The number of moles of

\mathrm{N}_2

in

28 \mathrm{~g}

is:

\text{Number of moles of } \mathrm{N}_2 = \frac{28 \mathrm{~g}}{28 \mathrm{~g/mol}} = 1 \mathrm{~mol}

Comparing these calculated values, we can see that

0.5 \mathrm{~mol}

of

\mathrm{N}_2

(Option A) is equal to

0.5 \mathrm{~mol}

of

\mathrm{H}_2

. Therefore, the correct answer is: Option A:

14 \mathrm{~g}

of

\mathrm{N}_2

.

Q7

On complete combustion, 0.3 g of an organic compound gave 0.2 g of CO

_2

and 0.1 g of H

_2

O. The percentage composition of carbon and hydrogen in the compound, respectively is:

A 4.07% and 15.02%

B 18.18% and 3.70%

C 15.02% and 4.07%

D 3.70% and 18.18%

Correct Answer

Option B

Solution

To determine the percentage composition of carbon and hydrogen in the organic compound, we need to calculate the amounts of carbon in CO

_2

and hydrogen in H

_2

O produced from the complete combustion of the compound. First, let's calculate the amount of carbon in CO

_2

: The molar mass of CO

_2

is 44 g/mol, and the molar mass of carbon (C) is 12 g/mol. Given that 0.2 g of CO

_2

is produced, we can use the following proportion to find the mass of carbon:

\frac{12 \text{ g (C)}}{44 \text{ g (CO}_2\text{)}} = \frac{x}{0.2 \text{ g (CO}_2\text{)}}

Solving for

x

(mass of carbon):

x = \frac{12 \times 0.2}{44} = \frac{2.4}{44} = 0.0545 \text{ g}

Next, let's calculate the amount of hydrogen in H

_2

O: The molar mass of H

_2

O is 18 g/mol, and the molar mass of hydrogen (H) in a single H

_2

O molecule is 2 g/mol. Given that 0.1 g of H

_2

O is produced, we can use the following proportion to find the mass of hydrogen:

\frac{2 \text{ g (H)}}{18 \text{ g (H}_2\text{O})} = \frac{y}{0.1 \text{ g (H}_2\text{O})}

Solving for

y

(mass of hydrogen):

y = \frac{2 \times 0.1}{18} = \frac{0.2}{18} = 0.0111 \text{ g}

Now, we can calculate the percentage composition of carbon and hydrogen in the organic compound, which has a total mass of 0.3 g.

Percentage of carbon:

\% \text{C} = \left( \frac{0.0545 \text{ g}}{0.3 \text{ g}} \right) \times 100 = 18.18\%

Percentage of hydrogen:

\% \text{H} = \left( \frac{0.0111 \text{ g}}{0.3 \text{ g}} \right) \times 100 = 3.70\%

Therefore, the correct answer is: Option B : 18.18% and 3.70%

Q8

1 gram of sodium hydroxide was treated with

25 \mathrm{~mL}

of

0.75 \mathrm{~M} \mathrm{~HCl}

solution, the mass of sodium hydroxide left unreacted is equal to

A 750 mg

B 250 mg

C Zero mg

D 200 mg

Correct Answer

Option B

Solution

To find the mass of sodium hydroxide (NaOH) left unreacted, we first need to determine the moles of NaOH and HCl originally present and compare them to see which one is in excess.

The molar mass of NaOH is approximately

40 \text{ g/mol}

, so the number of moles of NaOH in 1 gram can be calculated as follows:

\text{Moles of NaOH} = \frac{\text{Mass of NaOH}}{\text{Molar mass of NaOH}} = \frac{1 \text{ g}}{40 \text{ g/mol}} = 0.025 \text{ moles}

Next, we calculate the number of moles of HCl using its concentration and the volume of the solution.

Recall that concentration (Molarity, M) is defined as moles of solute per liter of solution.

Given that the concentration of HCl is

0.75 \text{ M}

and the volume of the solution is

25 \text{ mL}

or

0.025 \text{ L}

, we can find the moles of HCl:

\text{Moles of HCl} = \text{Concentration} \times \text{Volume in liters} = 0.75 \text{ M} \times 0.025 \text{ L} = 0.01875 \text{ moles}

Now, we compare the moles of NaOH and HCl. The stoichiometry of the reaction between NaOH and HCl is:

\text{NaOH} + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O}

Each mole of NaOH reacts with one mole of HCl.

Given that there are more moles of NaOH (0.025 moles) than HCl (0.01875 moles), HCl is the limiting reactant and will be completely consumed.

The excess NaOH can be calculated:

\text{Excess moles of NaOH} = \text{Moles of NaOH} - \text{Moles of HCl} = 0.025 \text{ moles} - 0.01875 \text{ moles} = 0.00625 \text{ moles}

To find the mass of the unreacted NaOH:

\text{Mass of unreacted NaOH} = \text{Moles of unreacted NaOH} \times \text{Molar mass of NaOH} = 0.00625 \text{ moles} \times 40 \text{ g/mol} = 0.25 \text{ g} = 250 \text{ mg}

Therefore, the mass of sodium hydroxide left unreacted is 250 mg , which corresponds to Option B.

Q9

The highest number of helium atoms is in

A

4 \mathrm{~mol}

of helium

B

\mathrm{4 u}

of helium

C

4 \mathrm{~g}

of helium

D

2.271098 \mathrm{~L}

of helium at STP

Correct Answer

Option A

Solution

To determine which option contains the highest number of helium atoms, we need to analyze each option based on the amount of helium it represents and apply Avogadro's Law as required.

Option A:

4 \, \text{mol}

of helium Using Avogadro's number, which is approximately

6.022 \times 10^{23}

atoms per mole, the number of helium atoms in 4 moles can be calculated as:

4 \, \text{mol} \times 6.022 \times 10^{23} \, \text{atoms/mol} = 24.088 \times 10^{23} \, \text{atoms}

Option B:

4 \, \text{u}

of helium The atomic mass of helium is approximately 4 u (atomic mass units). Therefore,

4 \, \text{u}

represents about 1 mole of helium atoms (since the molar mass of helium is approximately 4 g/mol, which equals 4 u).

Thus, this option represents:

1 \, \text{mol} \times 6.022 \times 10^{23} \, \text{atoms/mol} = 6.022 \times 10^{23} \, \text{atoms}

Option C:

4 \, \text{g}

of helium Similarly, as we've established that the molar mass of helium is 4 g/mol,

4 \, \text{g}

of helium equates exactly to:

1 \, \text{mol} \times 6.022 \times 10^{23} \, \text{atoms/mol} = 6.022 \times 10^{23} \, \text{atoms}

Option D:

2.271098 \, \text{L}

of helium at STP (Standard Temperature and Pressure) At STP, one mole of any ideal gas occupies 22.4 L.

Therefore, the amount of helium in moles for

2.271098 \, \text{L}

can be derived from:

\frac{2.271098 \, \text{L}}{22.4 \, \text{L/mol}} \approx 0.101 \, \text{mole}

Using this to find the number of atoms:

0.101 \, \text{mol} \times 6.022 \times 10^{23} \, \text{atoms/mol} \approx 6.082 \times 10^{22} \, \text{atoms}

Conclusion: Comparing the numbers: Option A:

24.088 \times 10^{23} \, \text{atoms}

Option B:

6.022 \times 10^{23} \, \text{atoms}

Option C:

6.022 \times 10^{23} \, \text{atoms}

Option D:

6.082 \times 10^{22} \, \text{atoms}

Option A clearly contains the highest number of helium atoms, which is

24.088 \times 10^{23} \, \text{atoms}.

Q10

A compound X contains

32 \%

of A,

20 \%

of B and remaining percentage of C. Then, the empirical formula of

\mathrm{X}

is : (Given atomic masses of A=64 ; B=40 ; C=32 u)

A

\mathrm{A}_2 \mathrm{BC}_2

B

\mathrm{ABC}_3

C

\mathrm{AB}_2 \mathrm{C}_2

D

\mathrm{ABC}_4

Correct Answer

Option B

Solution

<p> </p><table class=tg style=undefined;table-layout: fixed; width: 627px> <colgroup> <col style=width: 98px> <col style=width: 116px> <col style=width: 111px> <col style=width: 155px> <col style=width: 147px> </colgroup> <thead> <tr> <th class=tg-7btt>Element</th> <th class=tg-7btt>Mass percentage %</th> <th class=tg-7btt>No. of moles</th> <th class=tg-7btt>No. of moles/Smallest number</th> <th class=tg-7btt>Simplest whole number</th> </tr> </thead> <tbody> <tr> <td class=tg-c3ow>A</td> <td class=tg-c3ow>32%</td> <td class=tg-c3ow>

<br>\frac{32}{64}=\frac{1}{2}<br>

</td> <td class=tg-c3ow>

<br>\frac{1}{2} \times 2<br>

</td> <td class=tg-c3ow>= 1</td> </tr> <tr> <td class=tg-c3ow>B</td> <td class=tg-c3ow>20%</td> <td class=tg-c3ow>

<br>\frac{20}{40}=\frac{1}{2}<br>

</td> <td class=tg-c3ow>

<br>\frac{1}{2} \times 2<br>

</td> <td class=tg-c3ow>= 1</td> </tr> <tr> <td class=tg-c3ow>C</td> <td class=tg-c3ow>48%</td> <td class=tg-c3ow>

<br>\frac{48}{32}=\frac{3}{2}<br>

</td> <td class=tg-c3ow>

<br>\frac{3}{2} \times 2<br>

</td> <td class=tg-c3ow>= 3</td> </tr> </tbody> </table><p></p> <p>So, empirical formula of <img src=https://app-content.cdn.examgoal.net/fly/@width/image/1lvxdm5vz/13b0f776-95ca-45c6-8a99-73ffec243d46/6f29cdf0-0cfb-11ef-8086-578ef1b7ca01/file-1lvxdm5w0.png?

format=png data-orsrc=https://app-content.cdn.examgoal.net/image/1lvxdm5vz/13b0f776-95ca-45c6-8a99-73ffec243d46/6f29cdf0-0cfb-11ef-8086-578ef1b7ca01/file-1lvxdm5w0.png loading=lazy style=max-width:100%;height:auto;display:block;margin:0 auto;max-height:40vh;vertical-align:baseline alt=NEET 2024 Chemistry - Some Basic Concepts of Chemistry Question 8 English Explanation></p> <p>

\therefore \text{ The correct empirical formula of compound } \mathrm{X} \text{ is } \mathrm{ABC}_3

</p>