NCERT Solutions for Class 11 Science Chemistry Chapter 5 States Of Matter are provided here with simple step-by-step explanations. These solutions for States Of Matter are extremely popular among Class 11 Science students for Chemistry States Of Matter Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the NCERT Book of Class 11 Science Chemistry Chapter 5 are provided here for you for free. You will also love the ad-free experience on Meritnationâ€™s NCERT Solutions. All NCERT Solutions for class Class 11 Science Chemistry are prepared by experts and are 100% accurate.

#### Page No 152:

#### Question 5.1:

What
will be the minimum pressure required to compress 500 dm^{3}
of air at 1 bar to 200 dm^{3}
at 30°C?

#### Answer:

Given,

Initial
pressure, *p*_{1}
= 1 bar

Initial
volume, *V*_{1}
= 500 dm^{3}

Final
volume, *V*_{2}
= 200 dm^{3}

Since
the temperature remains constant, the final pressure (*p*_{2})
can be calculated using Boyle’s law.

According to Boyle’s law,

Therefore, the minimum pressure required is 2.5 bar.

#### Page No 153:

#### Question 5.2:

A vessel of 120 mL capacity contains a certain amount of gas at 35 °C and 1.2 bar pressure. The gas is transferred to another vessel of volume 180 mL at 35 °C. What would be its pressure?

#### Answer:

Given,

Initial
pressure, *p*_{1}
= 1.2 bar

Initial
volume, *V*_{1
}= 120 mL

Final
volume, *V*_{2}
= 180 mL

Since
the temperature remains constant, the final pressure (*p*_{2})
can be calculated using Boyle’s law.

According to Boyle’s law,

Therefore, the pressure would be 0.8 bar.

#### Page No 153:

#### Question 5.3:

Using
the equation of state *pV *=
*n*R*T*;
show that at a given temperature density
of a gas is proportional to gas pressure*p*.

#### Answer:

The equation of state is given by,

*pV* = *n*R*T* ……….. (i)

Where,

*p** → *Pressure of gas

*V* → Volume of gas

*n*→ Number of moles of gas

R → Gas constant

*T* → Temperature of gas

From equation (i) we have,

Replacing *n* with , we have

Where,

*m* → Mass of gas

*M* → Molar mass of gas

But, (*d* = density of gas)

Thus, from equation (ii), we have

Molar mass (*M*) of a gas is always constant and therefore, at constant temperature= constant.

Hence, at a given temperature, the density (*d*) of gas is proportional to its pressure (*p)*

#### Page No 153:

#### Question 5.4:

At 0°C, the density of a certain oxide of a gas at 2 bar is same as that of dinitrogen at 5 bar. What is the molecular mass of the oxide?

#### Answer:

Density
(d) of the substance at temperature (*T*)
can be given by the expression,

*d*
=

Now,
density of oxide (*d*_{1})
is given by,

Where,
*M*_{1}
and *p*_{1}
are the mass and pressure of the oxide respectively.

Density
of dinitrogen gas (*d*_{2})
is given by,

Where,
*M*_{2}
and *p*_{2}
are the mass and pressure of the oxide respectively.

According to the given question,

Molecular
mass of nitrogen, *M*_{2}
= 28 g/mol

Hence, the molecular mass of the oxide is 70 g/mol.

#### Page No 153:

#### Question 5.5:

Pressure of 1 g of an ideal gas A at 27 °C is found to be 2 bar. When 2 g of another ideal gas B is introduced in the same flask at same temperature the pressure becomes 3 bar. Find a relationship between their molecular masses.

#### Answer:

For ideal gas A, the ideal gas equation is given by,

Where,
*p*_{A}
and *n*_{A}
represent the pressure and number of moles of gas A.

For ideal gas B, the ideal gas equation is given by,

Where,
*p*_{B
}and *n*_{B}
represent the pressure and number of moles of gas B.

[*V*
and *T* are constants for gases A and B]

From equation (i), we have

From equation (ii), we have

Where,
M_{A }and M_{B} are the molecular masses of gases A
and B respectively.

Now, from equations (iii) and (iv), we have

Given,

(Since total pressure is 3 bar)

Substituting these values in equation (v), we have

Thus, a relationship between the molecular masses of A and B is given by

.

#### Page No 153:

#### Question 5.6:

The drain cleaner, Drainex contains small bits of aluminum which react with caustic soda to produce dihydrogen. What volume of dihydrogen at 20 °C and one bar will be released when 0.15g of aluminum reacts?

#### Answer:

The reaction of aluminium with caustic soda can be represented as:

At
STP (273.15 K and 1 atm), 54 g (2 ×
27 g) of Al gives 3 ×
22400 mL of H_{2..}

0.15 g Al gives
i.e.,
186.67 mL of H_{2.}

At STP,

Let
the volume of dihydrogen be
at *p*_{2}
= 0.987 atm (since 1 bar = 0.987 atm) and *T*_{2}
= 20°C
= (273.15 + 20) K = 293.15 K._{.}

Therefore, 203 mL of dihydrogen will be released.

#### Page No 153:

#### Question 5.7:

What will be the pressure exerted by a mixture of 3.2 g of methane and 4.4 g of carbon dioxide contained in a 9 dm^{3} flask at 27 °C ?

#### Answer:

It is known that,

For methane (CH_{4}),

For carbon dioxide (CO_{2}),

Total pressure exerted by the mixture can be obtained as:

Hence, the total pressure exerted by the mixture is 8.314** **× 10^{4} Pa.

#### Page No 153:

#### Question 5.8:

What will be the pressure of the gaseous mixture when 0.5 L of H_{2} at 0.8 bar and 2.0 L of dioxygen at 0.7 bar are introduced in a 1L vessel at 27°C?

#### Answer:

Let the partial pressure of H_{2} in the vessel be.

Now,

= ?

It is known that,

Now, let the partial pressure of O_{2} in the vessel be.

${}\phantom{\rule{0ex}{0ex}}{{\mathrm{p}}}_{1}{{\mathrm{V}}}_{1}{=}{{\mathrm{p}}}_{2}{{\mathrm{V}}}_{2}\phantom{\rule{0ex}{0ex}}{\Rightarrow}{{\mathrm{p}}}_{2}{=}\frac{{\mathrm{p}}_{1}{\mathrm{V}}_{1}}{{\mathrm{V}}_{2}}\phantom{\rule{0ex}{0ex}}{\Rightarrow}{{\mathrm{p}}}_{{\mathrm{O}}_{2}}{=}\frac{0.7\times 2.0}{1}{=}{}{1}{.}{4}{}{\mathrm{bar}}\phantom{\rule{0ex}{0ex}}$

Total pressure of the gas mixture in the vessel can be obtained as:

Hence, the total pressure of the gaseous mixture in the vessel is.

#### Page No 153:

#### Question 5.9:

Density of a gas is found to be 5.46 g/dm^{3} at 27 °C at 2 bar pressure. What will be its density at STP?

#### Answer:

Given,

The density (*d*_{2}) of the gas at STP can be calculated using the equation,

Hence, the density of the gas at STP will be 3 g dm^{–3}.

#### Page No 153:

#### Question 5.10:

34.05 mL of phosphorus vapour weighs 0.0625 g at 546 °C and 0.1 bar pressure. What is the molar mass of phosphorus?

#### Answer:

Given,

*p* = 0.1 bar

*V* = 34.05 mL = 34.05 × 10^{–3} L = 34.05 × 10^{–3} dm^{3}

R = 0.083 bar dm^{3} K^{–1} mol^{–1}

*T* = 546°C = (546 + 273) K = 819 K

The number of moles (*n*) can be calculated using the ideal gas equation as:

Therefore, molar mass of phosphorus = 1247.5 g mol^{–1}

Hence, the molar mass of phosphorus is 1247.5 g mol^{–1}.

#### Page No 153:

#### Question 5.11:

A student forgot to add the reaction mixture to the round bottomed flask at 27 °C but instead he/she placed the flask on the flame. After a lapse of time, he realized his mistake, and using a pyrometer he found the temperature of the flask was 477 °C. What fraction of air would have been expelled out?

#### Answer:

Let the volume of the round bottomed flask be *V.*

Then, the volume of air inside the flask at 27° C is *V.*

Now,

*V*_{1} = *V*

*T*_{1} = 27°C = 300 K

*V*_{2} =?

*T*_{2} = 477° C = 750 K

According to Charles’s law,

Therefore, volume of air expelled out = 2.5 *V* – *V* = 1.5 *V*

Hence, fraction of air expelled out

#### Page No 153:

#### Question 5.12:

Calculate
the temperature of 4.0 mol of a gas occupying 5 dm^{3
}at 3.32 bar.

(R
= 0.083 bar dm^{3}
K^{–1}
mol^{–1}).

#### Answer:

Given,

*n*
= 4.0 mol

*V*
= 5 dm^{3}

*p*
= 3.32 bar

R
= 0.083 bar dm^{3}
K^{–1}
mol^{–1}

The temperature (T) can be calculated using the ideal gas equation as:

Hence, the required temperature is 50 K.

#### Page No 153:

#### Question 5.13:

Calculate the total number of electrons present in 1.4 g of dinitrogen gas.

#### Answer:

Molar mass of dinitrogen (N_{2}) = 28 g mol^{–1}

Thus, 1.4 g of

Now, 1 molecule of contains 14 electrons.

Therefore, 3.01 × 10^{23} molecules of N_{2} contains = 1.4 × 3.01 × 10^{23}

= 4.214 × 10^{23} electrons

#### Page No 153:

#### Question 5.14:

How much time would it take to distribute one Avogadro number of wheat grains, if 10^{10} grains are distributed each second?

#### Answer:

Avogadro number = 6.02 × 10^{23}

Thus, time required

=

Hence, the time taken would be.

#### Page No 153:

#### Question 5.15:

Calculate the total pressure in a mixture of 8 g of dioxygen and 4 g of dihydrogen confined in a vessel of 1 dm^{3} at 27°C. R = 0.083 bar dm^{3} K^{–}^{1} mol^{–}^{1.}

#### Answer:

Given,

Mass
of dioxygen (O_{2}) = 8 g

Thus, number of moles of

Mass
of dihydrogen (H_{2}) = 4 g

Thus, number of moles of

Therefore, total number of moles in the mixture = 0.25 + 2 = 2.25 mole

Given,

*V*
= 1 dm^{3}

*n*
= 2.25 mol

R
= 0.083 bar dm^{3} K^{–1} mol^{–1}

*T*
= 27°C = 300 K

Total
pressure (*p*) can be calculated as:

*pV*
=* n*R*T*

Hence, the total pressure of the mixture is 56.025 bar.

#### Page No 153:

#### Question 5.16:

Pay load is defined as the difference between the mass of displaced air and the mass of the balloon. Calculate the pay load when a balloon of radius 10 m, mass 100 kg is filled with helium at 1.66 bar at 27°C. (Density of air = 1.2 kg m^{–}^{3} and R = 0.083 bar dm^{3} K^{–}^{1} mol^{–}^{1}).

#### Answer:

Given,

Radius of the balloon, *r* = 10 m

Volume of the balloon

Thus, the volume of the displaced air is 4190.5 m^{3}.

Given,

Density of air = 1.2 kg m^{–3}

Then, mass of displaced air = 4190.5 × 1.2 kg

= 5028.6 kg

Now, mass of helium (*m*) inside the balloon is given by,

Now, total mass of the balloon filled with helium = (100 + 1117.5) kg

= 1217.5 kg

Hence, pay load = (5028.6 – 1217.5) kg

= 3811.1 kg

Hence, the pay load of the balloon is 3811.1 kg.

#### Page No 153:

#### Question 5.17:

Calculate
the volume occupied by 8.8 g of CO_{2}
at 31.1°C and 1 bar pressure.

R
= 0.083 bar L K^{–1}
mol^{–1}.

#### Answer:

It is known that,

Here,

*m*
= 8.8 g

R
= 0.083 bar LK^{–1}
mol^{–1}

*T*
= 31.1°C
= 304.1 K

*M*
= 44 g

*p*
= 1 bar

Hence, the volume occupied is 5.05 L.

#### Page No 153:

#### Question 5.18:

2.9 g of a gas at 95 °C occupied the same volume as 0.184 g of dihydrogen at 17 °C, at the same pressure. What is the molar mass of the gas?

#### Answer:

Volume
(*V*)
occupied by dihydrogen is given by,

Let
M be the molar mass of the unknown gas. Volume (*V*)
occupied by the unknown gas can be calculated as:

According to the question,

Hence,
the molar mass of the gas is 40 g mol^{–1}.

#### Page No 153:

#### Question 5.19:

A mixture of dihydrogen and dioxygen at one bar pressure contains 20% by weight of dihydrogen. Calculate the partial pressure of dihydrogen.

#### Answer:

Let the weight of dihydrogen be 20 g and the weight of dioxygen be 80 g.

Then, the number of moles of dihydrogen, and the number of moles of dioxygen, .

Given,

Total pressure of the mixture, *p*_{total }= 1 bar

Then, partial pressure of dihydrogen,

Hence, the partial pressure of dihydrogen is.

#### Page No 153:

#### Question 5.20:

What
would be the SI unit for the quantity *pV*^{2}*T
*^{2}/*n?*

#### Answer:

The
SI unit for pressure, *p*
is Nm^{–2}.

The
SI unit for volume, *V*
is m^{3.}

The
SI unit for temperature, *T*
is K.

The
SI unit for the number of moles, *n *is
mol.

Therefore, the SI unit for quantity is given by,

#### Page No 153:

#### Question 5.21:

In terms of Charles’ law explain why –273°C is the lowest possible temperature.

#### Answer:

Charles’ law states that at constant pressure, the volume of a fixed mass of gas is directly proportional to its absolute temperature.

It was found that for all gases (at any given pressure), the plots of volume vs. temperature (in °C) is a straight line. If this line is extended to zero volume, then it intersects the temperature-axis at – 273°C. In other words, the volume of any gas at –273°C is zero. This is because all gases get liquefied before reaching a temperature of – 273°C. Hence, it can be concluded that – 273°C is the lowest possible temperature.

#### Page No 153:

#### Question 5.22:

Critical temperature for carbon dioxide and methane are 31.1 °C and –81.9 °C respectively. Which of these has stronger intermolecular forces and why?

#### Answer:

Higher
is the critical temperature of a gas, easier is its liquefaction.
This means that the intermolecular forces of attraction between the
molecules of a gas are directly proportional to its critical
temperature. Hence, intermolecular forces of attraction are stronger
in the case of CO_{2}.

#### Page No 153:

#### Question 5.23:

Explain the physical significance of Van der Waals parameters.

#### Answer:

**Physical
significance of ‘a’:**

‘a’ is a measure of the magnitude of intermolecular attractive forces within a gas.

**Physical
significance of ‘b’:**

‘b’ is a measure of the volume of a gas molecule.

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