States of Matter Test

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States of Matter Test - 45

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Weekly Quiz Competition

Question 1
1 / -0

A relation between vapour pressure and the temperature is known as

A
Ideal gas equation

B
Boltzman equation

C
Clausious equation

D
Clausius-Clapeyron equation

Solution

Option D is correct. A relation between vapour pressure and the temperature is known as the Clausius-Clapeyron equation.
Question 2
1 / -0

A container contains $$O_{2}$$ and $$N_{2}$$ in equal molar concentration at same temperature, what is the $$CORRECT$$ statement about the average molar kinetic energy of the two gasses?

A
Depends upon volume

B
$$KE_{N2}=KE_{O2}$$

C
$$KE_{N2}> KE_{O2}$$

D
$$KE_{N2}< KE_{O2}$$

Solution

Molar kinetic energy$$=KE= \dfrac 32 RT$$

It is dependent on temperature.

Average kinetic is directly proportional to temperature.

Hence average KE is same as temperature is same.

So the correct answer is(B) $$KE_{N_2}=KE_{O_2}$$
Question 3
1 / -0

Lowering in vapour pressure is highest for:

A
$$0.2m$$ urea

B
$$0.1m $$ glucose

C
$$0.1m \, MgSO_4$$

D
$$0.1m \, BaCl_2$$

Solution

Lowering in vapour pressure is directly proportional
to molality which in turn is directly proportional
to n factor (no. of ions it gives)
The concentration of urea, i.e, $$0.2 \ M$$ is much
more than glucose, $$MgSO_{4}$$ & $$BaCl_{2}$$
Therefore urea will have greater molality &
therefore it will have highest vapour pressure.
Ans is A) $$0.2 M$$ urea
Question 4
1 / -0

A mixture contains 16 g of oxygen ,28g of nitrogen and 8g of methane.Total pressure of the mixture is 740mm. What is the partial pressure of nitrogen in mm?

A
185

B
370

C
555

D
740

Solution

As no of moles of $$O_2=16/32=0.5$$ moles
no of moles of $$N_2=28/28=1$$ moles
no of moles of $$CH_4=8/16=0.5$$ moles
Total pressure=740mm
Partial pressure=total pressure$$\times$$ mole fraction of $$N_2$$
Mole fraction of $$N_2$$ = No of moles of $$N_2$$ /total no of moles
$$=1/0.5+0.5+1=1/2$$
then partial pressure $$=1/2\times740$$
$$=370\ mm$$
Question 5
1 / -0

if the mass ratio of hydrogen gas is to oxygen gas in a mixture is 1:2 then the ratio of their number of molecules is:

A
4:1

B
1:2

C
8:1

D
1:4

Solution

No of molecules = $$\dfrac{Weight}{Gram\space Molecular\space Weight} \times {N}_{A}$$
Molecular wt of hydrogen $${H}_{2}$$ = 2
Molecular wt of oxygen $${O}_{2}$$ = 32
So Ratio of number of molecules = $$\dfrac{1}{2} : \dfrac{2}{32}$$ = $$8 : 1$$

Option $$C$$ is correct.
Question 6
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A container of volume $$40$$ liters consist of some gas at a pressure of 2 atm and temperature of $$300 K$$. It is heated to $$400 K$$ such that half of the gas escape and volume changes to 60 liters. The new pressure of the gas at the final condition will be:

A
$$2 atm$$

B
$$1 atm$$

C
$$\dfrac{8}{9}$$ atm

D
$$\dfrac{9}{8}$$ atm

Solution
Question 7
1 / -0

In an ideal solution of non-volatile solute B in solvent A in 2: 5 molar ratio has vapour pressure 250 mm. If another solution in ratio 3: 4 prepared then vapour pressure above this solution is:

A
200 mm

B
250 mm

C
350 mm

D
400 mm

Solution

Given molar ratio$$=2:5$$ v.p $$=250\ mm\left(\text{one solute}\right)$$

Moral ratio $$=3:4\quad v.p=? \quad\left(\text{another solute}\right)$$

Mole fraction of solven $$=\dfrac{5}{2+5}=\dfrac{5}{4}=0.71\quad \left({1}^{st}{\text{solute}}\right)$$

According to Raoult's law;

$$p$$ (solution)$$\times$$ mole fraction of solven$$\Rightarrow250=p\left(\text{pure solvent}\right)0.71$$

$$p\left(\text{pure solvent}\right)= \dfrac{250}{0.71}=352.11\ mm$$

mole fraction of solvent $$=\dfrac{4}{3+4}=\dfrac{4}{7}=0.57$$

$$p\left(\text{solution}\right)=352.11\times 0.57=200.7\approx 200\ mm$$

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

A sample of 16g charcoal was brought into contact with $$ C{ H}_{ 4}$$ gas contained in a vessel of 1 litre at $$ 2{ 7 }^{0 }C $$ . The pressure of gas was found to fall from 760 to 608 torr. The density of charcoal sample is 1.6g/ $$ c{ m}^{ 3 } $$. What is the volume of the $$ C{ H}_{ 4}$$ gas adsorbed per gram of the adsorbent at 608 torr and $$ 2{ 7 }^{0 }C $$ ?

A
125 mL/g

B
16.25 mL/g

C
26 mL/g

D
None of these

Solution

Final volume of gas, at $$608$$ torr pressure
$$V_{2} = \dfrac{P_1V_1}{P_2} $$

$$ = \dfrac{760 \times 1}{608} $$

$$ \Rightarrow $$ $$1.25$$ or $$V_2 = 1250\,mL $$

Volume occupied by gas = Volume of vessel - Volume occupied by charcoal

$$ = 1000 - \dfrac{16}{1.6} = 990 \,mL $$

Difference of volume is due to adsorption of gas by charcoal.
$$ \therefore $$ Volume of gas adsorbed by charcoal
$$ = 1250 - 990 $$
$$ \Rightarrow 260\,mL$$
Volume of the gas adsorbed per gram of charcoal
$$ = \dfrac{260}{16} = 16.25\,mL / g $$ at $$608$$ torr and $$27^{\circ}C $$.
Question 9
1 / -0

Equal weight of $${CH}_{4}$$ and $${H}_{2}$$ are mixed in a container at $$25^o C$$, the fraction of total pressure exerted by hydrogen is :

A
$$\cfrac { 1 }{ 2 }$$

B
$$\cfrac { 1 }{ 3 }$$

C
$$\cfrac { 1 }{ 9 }$$

D
$$\cfrac { 8 }{ 9 }$$

Solution

Let the weight of methane and hydrogen added be x grams.

Then, moles of methane$$ = \dfrac{x}{{16}}$$

And, moles of hydrogen$$ = \dfrac{x}{2}$$

Now , we know that $$PV = nRT$$.(where P is pressure, n is the number of moles, V is the volume, T is the temperature, and R is the gas constant.)

In this case, V, R and T are constant. Hence,$$P\propto n$$$$ \Rightarrow {P_{Total}}= \dfrac{x}{{16}} + \dfrac{x}{2} = \dfrac{9}{{16}}x$$ $$ \to (1)$$

Also, $${P_{Hydrogen}}\propto \dfrac{x}{2}$$ $$ \to (2)$$

Now, dividing equation (2) by (1), we get fraction of total pressure exerted by hydrogen$$ = \dfrac{{\dfrac{x}{2}}}{{\dfrac{{9x}}{{16}}}} = \dfrac{8}{9}$$, which is the required answer.
Question 10
1 / -0

The total kinetic energy (in joules) of the molecules in $$8 \ g$$ of methane at $$27^oC$$ is:

A
$$374100 \ J$$

B
$$935.3 \ J$$

C
$$1870.65 \ J$$

D
$$700 \ J$$

Solution

$$KE=\cfrac{3}{2} nRT$$
8gm of $$CH_{4}=\cfrac{8}{16}=0.5 \,\,mol$$
$$\therefore$$ KE of 8gm of $$CH_{4}$$ at $$27^{\circ}C=\cfrac{3}{2}\times 0.5\times 8.314 \times 300=1870.65 \,\,J$$

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Re-Attempt Weekly Quiz Competition

States of Matter Test - 45

Result

States of Matter Test - 45

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SHARING IS CARING

Question 1

Ideal gas equation

Boltzman equation

Clausious equation

Clausius-Clapeyron equation

Solution

Question 2

Depends upon volume

$$KE_{N2}=KE_{O2}$$

$$KE_{N2}> KE_{O2}$$

$$KE_{N2}< KE_{O2}$$

Solution

Question 3

$$0.2m$$ urea

$$0.1m $$ glucose

$$0.1m \, MgSO_4$$

$$0.1m \, BaCl_2$$

Solution

Question 4

185

370

555

740

Solution

Question 5

4:1

1:2

8:1

1:4

Solution

Question 6

$$2 atm$$

$$1 atm$$

$$\dfrac{8}{9}$$ atm

$$\dfrac{9}{8}$$ atm

Solution

Question 7

200 mm

250 mm

350 mm

400 mm

Solution

Question 8

125 mL/g

16.25 mL/g

26 mL/g

None of these

Solution

Question 9

$$\cfrac { 1 }{ 2 }$$

$$\cfrac { 1 }{ 3 }$$

$$\cfrac { 1 }{ 9 }$$

$$\cfrac { 8 }{ 9 }$$

Solution

Question 10

$$374100 \ J$$

$$935.3 \ J$$

$$1870.65 \ J$$

$$700 \ J$$

Solution

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