States of Matter Test

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States of Matter Test - 32

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Weekly Quiz Competition

Question 1
1 / -0

At what relative humidity will $$Na_2SO_4$$ be deliquescent (absorb moisture) when exposed to the air at $$0^o$$C?
Given: $$Na_2SO_4\cdot 10H_2O(s)\rightleftharpoons Na_2SO_4(s)+10H_2O(g); K_p=4.08\times 10^{-25}$$ and vapour pressure of water $$0^o$$C $$=4.58$$ Torr.

A
Below $$50.5\%$$ but Above $$30.5\%$$

B
Above $$50.5\%$$

C
Above $$30.5\%$$

D
Below $$30.5\%$$

Solution
Question 2
1 / -0

Read the Paragraph carefully and answer the following questions:
100 g of an ideal $$\left( {mol.wt. = 40} \right)$$ gas is present in a cylinder at $${27^0}c\;and\,2atm$$ pressure. During transportation, cylinder fell and a dent was developed in the cylinder. The valve attached to cylinder cannot keep the pressure greater than $$2atm$$ and therefore $$'10g'$$ of gas leaked-out through cylinder.
The volume of the cylinder before a dent was:

A
$$10.79litre$$

B
$$20.79litre$$

C
$$30.79litre$$

D
$$40.79litre$$

Solution

Initial volume can be calculated:
$$PV=nRT$$
$$V_1=(\dfrac{100}{40})(\dfrac{0.0821 \times 300}{2}) = 30.79 dm^3$$
Question 3
1 / -0

The vapour pressure of two pure liquids $$(A)$$ and $$(B)$$ are $$100$$ and $$80\ torr$$ respectively. The total vapour pressure of solution obtained by mixing $$2$$ moles of $$(A)$$ and $$3$$ moles of $$(B)$$ would be :-

A
$$120\ torr$$

B
$$36\ torr$$

C
$$88\ torr$$

D
$$180\ torr$$

Solution

Total vapour pressure $$P={ P }_{ A }+{ P }_{ B }$$
$${ P }_{ A }=100\times \dfrac { 2 }{ 2+3 } =100\times \dfrac { 2 }{ 5 } =40$$ torr
$${ P }_{ B }=80\times \dfrac { 3 }{ 2+3 } =80\times \dfrac { 3 }{ 5 } =48$$ torr
$${ P }_{ Total }={ P }_{ A }+{ P }_{ B }=40+48=88$$ torr.
Question 4
1 / -0

The density of a gas is equal to:
($$P =$$ pressure, $$V=$$ volume, $$T=$$ temperature, $$R=$$ gas constant, $$n=$$ number of moles and $$M=$$ molecular weight).

A
$$nP$$

B
$$\dfrac{PM_w}{RT}$$

C
$$\dfrac{P}{RT}$$

D
$$\dfrac{M_w}{V}$$

Solution

According to ideal gas equation for monoatomic gas,
$$PV=RT$$
Volume $$=\dfrac { Mass/Molecular\quad weight }{ Density(D) } $$
$$\dfrac { P{ M }_{ w } }{ D } =RT$$
$$D=\dfrac { P{ M }_{ w } }{ RT } $$
Question 5
1 / -0

A closed bulb capacity 200ml containing $${ CH }_{ 4 }$$, $${ H }_{ 2 }$$ and He at 300K. The ratio of partial pressure of $${ CH }_{ 4 }$$, $${ H }_{ 2 }$$ and He, respectively is 2:3:5. Calculate the ratio of their weights present in the container.

A
2:4:6

B
16:8:100

C
16:3:10

D
4:9:25

Solution

Partial pressure $$=$$ mole fraction $$\times $$ total pressure
$$\therefore $$ Mass are also in $$2:3:5$$

$$\therefore $$ Ratio of mass $$=2\times 16:3\times 2:5\times 4$$
$$=16:3:10$$
Question 6
1 / -0

Ideal gas equation is also called equation of states because:

A
it depends on states of matter

B
it is relation between four variables and describes the state of any gas

C
it is combination of various gas laws and any variable can be calculated

D
none of these

Solution

Assuming an equilibrium state, the three properties needed to completely define the state of a system are pressure ($$P$$), volume ($$V$$), and temperature ($$T$$). Hence, the definition: An Equation of State (EOS) is a semi-empirical functional relationship between pressure, volume and temperature of a pure substance. Ideal gas equation is also called equation of states because it is relation between four variables and describes the state of any gas. Hence option $$B$$ is correct answer.
Question 7
1 / -0

What would be the vapour pressure of 0.5 molal solution of non volatile solute in benzene at $$30^oC$$ ? The vapour pressure of pure benzene at $$30^oC$$ is 119.6 torr.

A
119.6 torr

B
121.23 torr

C
118.52 torr

D
None of these

Solution

$$\dfrac{\triangle p}{p^0}=x_2$$
$$\dfrac{\triangle p}{p^0}=\dfrac{n_2}{n_1}$$
$$=\dfrac{m_2\times M_1}{M_2\times m_1}$$
$$=\dfrac{m\times m_1}{1000}$$
$$\dfrac{\triangle p}{p^0}=\dfrac{0.5\times 18}{1000}$$
$$=0.009$$
$$\dfrac{p^0-p}{p^0}=0.009$$
$$[m=molarity=\dfrac{m_2}{M_2}\times \dfrac{1000}{m_1}$$
$$\dfrac{119.6-p}{119.6}=009$$
$$19.6-p=1.0764$$
$$\boxed{p=118.52}torr$$
Question 8
1 / -0

An evacuated glass vessel weights 50.0 g when empty. 148.0 gm when filled with a liquid of density 0.98 g/mL and 50.5 g when filled with an ideal gas at 760 mm Hg at 300 K. Determine the molecular weight of the gas.

A
125

B
141

C
152

D
none of these

Solution

mass of containers $$=50g$$
$$\therefore$$ mass of liquid $$=(148-50)$$
$$=98g$$
$$density=0.98g/mL$$
$$V=\dfrac{m}{d}$$
$$=\dfrac{98}{0.98}\times 100$$
$$=100mL$$
$$=0.1mL$$
mass of gas $$=(50.5-50)g$$
$$=0.5g$$
$$\Rightarrow P=760mm$$ of $$Hg=1atm$$
$$T=300K$$
$$\therefore PV=nRT$$
$$1\times 0.1=n\times \dfrac{1}{12}\times 300$$
$$n\times 25=0.1$$
$$n=\dfrac{0.1}{250}$$
$$n=0.004$$
$$\therefore $$ mass of $$0.004mole=0.5g$$
$$\therefore $$ mass of $$0.004mole=\dfrac{0.5\times 1000}{4}$$
$$=125g$$
Question 9
1 / -0

The vapour pressure of pure water is 760 mm at $$25^{ 0 }C$$.The vapour pressure of solution containing 1(m) solution of glucose will be:

A
761.0 mm

B
746.5 mm

C
734.5 atm

D
746.5 Pa

Solution

Vapor pressure
$$\dfrac{P^{o}- P}{P^{o}}= \dfrac{W_{2}}{M_{2}} \times \dfrac{M_{1}}{W_{1}}$$

$$P^{o}= 760\ mm$$

$$M_{1} = 18$$

Molality $$(m)= 1m= \dfrac{mole\ of\ solute}{(weight)\ kg\ of\ solvent }= \dfrac{W_{2}/M_{2}}{W_{i} m \ gm}$$

$$\dfrac{760-P}{760} = 1 \times 10^{-3} \times 18$$

$$P= 746.5\ mm$$

Hence, the correct option is $$B$$
Question 10
1 / -0

A mixture of $$He$$ and $$SO_2$$ at one bar pressure contains $$20\%$$ by weight of $$He$$. Partial pressure of $$He$$ be:

A
0.2 bar

B
0.4 bar

C
0.6 bar

D
0.8 bar

Solution

Mixture contains 20g $$He$$ and 80g $$S{O}_{4}$$
No of moles of $$He$$ = 20/4 = 5 mol
No of moles of $$S{O}_{4}$$ = 80/64 = 1.25 mol
Mole Fraction of $$He$$ = 5/6.25 = 0.8
And Partial Pressure of $$He$$= 0.8 x 1 bar
= 0.8 bar

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States of Matter Test - 32

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States of Matter Test - 32

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Question 1

Below $$50.5\%$$ but Above $$30.5\%$$

Above $$50.5\%$$

Above $$30.5\%$$

Below $$30.5\%$$

Solution

Question 2

$$10.79litre$$

$$20.79litre$$

$$30.79litre$$

$$40.79litre$$

Solution

Question 3

$$120\ torr$$

$$36\ torr$$

$$88\ torr$$

$$180\ torr$$

Solution

Question 4

$$nP$$

$$\dfrac{PM_w}{RT}$$

$$\dfrac{P}{RT}$$

$$\dfrac{M_w}{V}$$

Solution

Question 5

2:4:6

16:8:100

16:3:10

4:9:25

Solution

Question 6

it depends on states of matter

it is relation between four variables and describes the state of any gas

it is combination of various gas laws and any variable can be calculated

none of these

Solution

Question 7

119.6 torr

121.23 torr

118.52 torr

None of these

Solution

Question 8

125

141

152

none of these

Solution

Question 9

761.0 mm

746.5 mm

734.5 atm

746.5 Pa

Solution

Question 10

0.2 bar

0.4 bar

0.6 bar

0.8 bar

Solution

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