States of Matter Test

Result

States of Matter Test - 42

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

$$K_p$$ value for $$2SO_{2(g)} + O_{2(g)} \rightleftharpoons 2SO_{3(g)}$$ is $$5.0 atm^{-1}$$. What is the equilibrium partial pressure of $$O_2$$ if the equilibrium pressures of $$SO_2$$ and $$SO_3$$ are equal?

A
$$0.2$$

B
$$0.3$$

C
$$0.4$$

D
$$0.1$$

Solution

For the given reaction,
$$2SO_2+O_2 \rightarrow 2SO_3$$
$$K_p$$ can be calculated as,
$$K_p=\dfrac{[P_{SO_3}^2]}{[P_{SO_2}]^2\times [P_{O_2}]}$$
As given,
$$[P_{SO_3}]=[P_{SO_2}]$$
So,
$$5=\dfrac{1}{[P_{O_2}]}$$
$$\implies [P_{O_2}]=\dfrac{1}{5}=0.2atm$$
Question 2
1 / -0

A mixture of ethyl alcohol and propyl alcohol has vapour pressure of $$290mm$$ at $$300K$$. The vapour pressure of propyl alcohol is $$200mm$$. If mole fraction of ethyl alcohol is $$0.6$$ its vapour presssure at same temperature will be:-

A
$$360$$

B
$$350$$

C
$$300$$

D
$$700$$

Solution

as $$P_{sol} = X_{sol} . P^o_{sol}$$    $$X \rightarrow $$ mole fraction
                                    $$P^o \rightarrow $$ pure vapour pressure of solution
$$290 mm =X_{ethanal} . P^o_{ethanal}   + X_{propanal} . P^o_{propanal}$$
$$290 mm =0.6 \times P^o_{ethanal}   + (1-0.6) \times 200 mm$$
on solving
$$P^o_{ethanal} =350$$ $$nm$$
Question 3
1 / -0

The value of the molar gas constant is:

A
$$8.3145\times 10^3 \ J\ mol^{-1}K^{-1}$$

B
$$1.987 \ cal\ mol^{-1} K^{-1}$$

C
$$0.083145\times 10^3dm^3 bar\ mol^{-1}K^{-1}$$

D
$$0.983145\ dm^3\ bar\ mol^{-1}K^{-1}$$

Solution

The gas constant is also known as the molar, universal, or ideal gas constant, denoted by the symbol R and is equivalent to the Boltzmann constant, but expressed in units of energy per temperature increment per mole.
Its value is $$8.31446J.K^{−1}.mol^{−1}$$
We know 1 joule = 4.182 cal
so its value becomes $$8.31446 \times 4.182 cal.K^{−1}.mol^{−1}$$ = $$1.987 cal.mol^{−1}.K^{−1}$$
Question 4
1 / -0

The amount of solute (mol mass=$$60$$) that must be added to $$180g$$ of water so that the vapour pressure of water is lowered by $$10 \%$$ is?

A
$$30$$g

B
$$60$$g

C
$$120$$g

D
$$12$$g

Solution
Question 5
1 / -0

1litre of a gas weights 2g at 300K and 1atm pressure.If the pressure is made 0.75atm at which temperature will 1L of same gas weighs 1g ?

A
600k

B
900K

C
800K

D
450K

Solution

$$P_1$$ = 1 atm ; $$W_1$$ = 2g

$$P_2$$ = 0.75 atm ; $$W_2$$ = 1 g

$$T_1$$ = 300 K ; $$V_1$$ = 1 L

$$T_2$$ = ? ; $$V_2$$ = 1 L

So we know that,

$$P_1V_1 / P_2V_2 = W_1T_1 / W_2T_2$$

Substituting in the above formula we get,

$$= 1 \times 1 / 0.75 \times 1 = 2 \times 300 / 1 \times T_2$$

$$= 1 / 0.75 = 600 / T_2$$

Cross Multiplying we get,

$$= T_2 \times 1 = 600 \times 0.75$$

$$= T_2 = 450$$ K
Question 6
1 / -0

At what temperature, the sample of neon gas would be heated to double its pressure initial volume of gas is reduced by 15% at $$75^{0}C $$?

A
$$592^{0}C$$

B
$$319^{0}C$$

C
$$789^{0}C$$

D
None

Solution

$$T_1= 273 + 75K = 348$$

$$\dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}$$

$$\dfrac{P_1V_1}{348}= \dfrac{2P_1\times 85V_1}{T_2\times 100}=591.6K$$

Temperature in $$C^0 = 591.6-273=31 9C^0$$
'
Question 7
1 / -0

How much pressure will be felt by a gas ballon which at 100 m depth in a sea form sea level?

A
$$1081320$$ Pa

B
$$98$$ Pa

C
$$9.8 \times 10^5$$ Pa

D
$$1.58 \times 10^6$$ Pa

Solution

$$P=P_O+\rho_gh$$

$$P_O=$$ atmospheric pressure$$=1.013 \times 10^5 Pa$$

$$\rho=$$ density of $$H_2O= 1000 kg/m^3, g=gravity= 9.8 m/s^2$$

$$P=$$ pressure experienced by gas balloon

$$h=$$ depth of the balloon from sea level= $$100m$$

$$P=(1.0132 \times 10^5)+(1000)(9.8)(100)$$

$$P=1081320 Pa$$

So, the correct option is $$A$$
Question 8
1 / -0

$$Mg + H_2SO_4 \rightarrow MgSO_4 + H_2$$
In this reaction, which mass of magnesium sulphate is formed when 6g of magnesium react with excess sulphuric acid?

A
$$8$$

B
$$24$$

C
$$30$$

D
$$60$$

Solution

Solution:- (C) $$30$$
As we know that,
$$\text{No. of moles} = \cfrac{\text{wt.}}{\text{mol. wt.}}$$
Now,
Wt. of $$Mg = 6 \; g$$
Mol. wt. of $$Mg = 24 \; g$$
No. of moles of $$Mg$$ in $$6 \; g = \cfrac{6}{24} = 0.25 \text{ moles}$$
$$Mg + {H}_{2}S{O}_{4} \longrightarrow MgS{O}_{4} + {H}_{2}$$
from the above reaction,
$$1$$ mole of magnesium sulphate is formed when $$1$$ mole of magnesium reacts.
Therefore,
$$0.25$$ mole of magnesium sulphate is formed when $$0.25$$ mole of magnesium reacts.
Now,
Mol. wt. of $$MgS{O}_{4} = 120 \; g$$
Therefore,
Wt. of $$0.25$$ moles of $$MgS{O}_{4} = 0.25 \times 120 = 30 \; g$$
Hence $$30 \; g$$ of magnesium sulphate is formd when $$6 \; g$$ of magnesium reacts with excess of $${H}_{2}S{O}_{4}$$.
Question 9
1 / -0

Which is not true in case of an ideal gas?

A
It cannot be converted into a liquid

B
There is no interaction between the molecules

C
All molecules of the gas move with same speed

D
At a given temperature, $$PV$$ is proportional to the amount of the gas

Solution

Solution:- (C) All molecules of the gas move with same speed
Molecules in an ideal gas move with different speed.
Question 10
1 / -0

The vapour pressure of water is $$12.3k \, Pa$$ at $$300K$$. Calculate the vapour pressure $$1$$ molal solution of a non-volatile solute in it.

A
$$12.08 \ kPa$$

B
$$1.208 \ Ppa$$

C
$$2.4 \ kPa$$

D
$$0.4 kPa$$

Solution

$$1$$ molal solution means 1 mol of the solute is present in $$1000 g$$ of the solvent (water).
The molar mass of water $$= 18 g$$ $$mol^{- 1}$$
Number of moles present in $$1000$$ g of water $$= 1000/18$$
$$= 55.56 \ mol$$
Therefore, the mole fraction of the solute in the solution is
$$x_2 = 1 / (1+55.56) $$
$$= 0.0177$$
It is given that,
Vapour pressure of water, $$P^0_1$$
$$= 12.3 kPa$$
Applying the relation, $$(P^0_1 - P_1) / P^0_1 = X_2$$
$$= (12.3 - P_1) / 12.3 $$
$$=0.0177$$
$$= 12.3 - P_1 $$
$$= 0.2177$$
$$= P_1 $$
$$= 12.0823$$
$$= 12.08 kPa$$

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

States of Matter Test - 42

Result

States of Matter Test - 42

Score

-

Rank

-

SHARING IS CARING

Question 1

$$0.2$$

$$0.3$$

$$0.4$$

$$0.1$$

Solution

Question 2

$$360$$

$$350$$

$$300$$

$$700$$

Solution

Question 3

$$8.3145\times 10^3 \ J\ mol^{-1}K^{-1}$$

$$1.987 \ cal\ mol^{-1} K^{-1}$$

$$0.083145\times 10^3dm^3 bar\ mol^{-1}K^{-1}$$

$$0.983145\ dm^3\ bar\ mol^{-1}K^{-1}$$

Solution

Question 4

$$30$$g

$$60$$g

$$120$$g

$$12$$g

Solution

Question 5

600k

900K

800K

450K

Solution

Question 6

$$592^{0}C$$

$$319^{0}C$$

$$789^{0}C$$

None

Solution

Question 7

$$1081320$$ Pa

$$98$$ Pa

$$9.8 \times 10^5$$ Pa

$$1.58 \times 10^6$$ Pa

Solution

Question 8

$$8$$

$$24$$

$$30$$

$$60$$

Solution

Question 9

It cannot be converted into a liquid

There is no interaction between the molecules

All molecules of the gas move with same speed

At a given temperature, $$PV$$ is proportional to the amount of the gas

Solution

Question 10

$$12.08 \ kPa$$

$$1.208 \ Ppa$$

$$2.4 \ kPa$$

$$0.4 kPa$$

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.