States of Matter Test

Result

States of Matter Test - 68

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Weekly Quiz Competition

Question 1
1 / -0

18 g of glucose $${ (C }_{ 6 }{ H }_{ 12 }{ O }_{ 6 })$$ is added to 178.2 g of water. The vapour pressure of water for this aqueous solution at $${ 100 }^{ \circ }C$$ is ?

A
7.60 Torr

B
76.00 Torr

C
752.40 Torr

D
759.00 Torr

Solution
Question 2
1 / -0

Gay Lussac's law is not valid for which chemical reaction?

A
$$2NO(g)+O_2(g)\longrightarrow 2NO_2(g)$$

B
$$2HI(g)\longrightarrow H_2(g)+I_2(g)$$

C
$$4B(s)+3O_2(g)\longrightarrow 2B_2O_3(s)$$

D
$$2SO_2(g)+O_2(g)\longrightarrow 2SO_3(g)$$
Question 3
1 / -0

A mixture of 50.ml of $$N{H_3}$$ and 60.0 ml of $${O_2}$$ gas react as
$$4N{H_3}(g) + 5{O_2}(g)\xrightarrow{{}}4NO + 6{H_2}O(g)$$
If all the gases are at the same temperature and the reaction continues until one of the gases is completely consumed, what volume of water vapour is produced?

A
48 mL

B
60.0mL

C
72 mL

D
75.0 mL

Solution
Question 4
1 / -0

The equilibrium constant for the following equilibrium is given at $$0^0C$$:

$$Na_2HPO_4.12H_2O(s)\rightarrow Na_2HPO_4.7H_2O(s)+5H_2O(g); K_p=31.25 \times 10^{-13}$$.

The vapor pressure of water is:

A
$$2 \times 10^{-4}\,atm$$

B
$$5 \times 10^{-4}\,atm$$

C
$$5 \times 10^{-2}\,atm$$

D
$$5 \times 10^{-3}\,atm$$
Question 5
1 / -0

Calculate the mass of a non - volatile solute (molar mass $$40\,\,g mol^{-1}$$) which should be dissolved in $$114g$$ octane to reduce its vapour pressure to $$80\%$$

A
$$2g$$

B
$$4g$$

C
$$8g$$

D
$$10g$$

Solution
Question 6
1 / -0

At $${40^ \circ }C$$ the vapour pressure ( in torr ) of a mixture of methyl alcohol and ethyl alcohol is represented by $$P = 199x + 135$$
(where $$x$$ is the mole fraction of methyl alcohol)
What are the vapour pressures of pure methyl alcohol and pure ethyl alcohol at $${40^ \circ }C$$ ?

A
$$135$$ and $$254$$ torr

B
$$119$$ and $$135$$ tor

C
$$119$$ and $$254$$ torr

D
$$334$$ and $$135$$ torr

Solution

Correct option is D
Question 7
1 / -0

Calculate the vapour pressure of aqueous 0.1 m glucose solution at 300 K temperature, the vapour pressure of water is 0.03 bar at 300 K temperature.

A
0.5 bar

B
0.29 bar

C
0.3 bar

D
0.03 bar

Solution
Question 8
1 / -0

$$5.0\ L$$ water placed in a closed room of volume $$2.5\times 10^{4}L$$ having temperature $$300\ K$$. If vapour pressure of water is $$27.0\ mm$$ and density is $$0.990\ g/cm^{3}$$ at this temperature, how much water is left in liquid state?

A
$$3.444\ L$$

B
$$4.344\ L$$

C
$$4.798\ L$$

D
$$1.212\ L$$
Question 9
1 / -0

At a certain temperature pure liquid A and liquid B have vapour pressures 10 torr and 37 torr respectively. For a certain ideal solution of A and B, the vapour in equilibrium with the liquid has the components A and B in the partial pressure ratio $$P_A : P_B=1 : 7 $$. What is the mole fraction of A in the solution?

A
0.346

B
0.654

C
0.5

D
None of these

Solution
Question 10
1 / -0

When the pressure on a gas is decreased to $$1/4$$ and the absolute temperature is increased four-fold the volume of the gas is:

A
Increases by $$16$$ times

B
Decreases to $$1/16$$

C
Increases by $$8$$ times

D
Remain the same

Solution

Hint: The gas which obeys all gas laws is called the ideal gas

Explanation:
Ideal gas law is

$$PV=nRT$$
Rearranging equation we get;

$$V = \dfrac{{nRT}}{P}$$

Where $$P, V $$ are respectively pressure and volume of the gas.

$$n$$ is the number of moles

$$T$$ is the temperature of the gas.

$$R$$ is ridberg constant.

The pressure on a gas is decreased to $$\dfrac{1}{4}$$ and the absolute temperature is increased four-fold then the volume of the gas is-
Let, $$P_1, V_1$$ and $$T_1$$ be the pressure, volume, and temperature of the initial stage and $$P_2, V_2$$ and $$T_2$$ be the pressure, volume, and temperature of the final stage.
Given that;

$$P_1=\dfrac{P_2}{4} $$ $$- - (i)$$
$$ P_2 = 4 \times P_1$$

$$T_1=4T_2$$ $$ - - (ii)$$
$$ \dfrac {P_1V_1}{T_1} = \dfrac {P_2V_2} {T_2}$$ - ( From Ideal gas as $$n$$ and $$R$$ are constant )
Putting values of $$P_2$$ and $$T_1$$ $$(i)$$ and $$(ii)$$ in above equation we get;
$$ \dfrac {P_1V_1}{4T_2} = \dfrac {4P_1V_2} {T_2}$$

Rearranging equation we get;
$$V_1 = 16 V_2$$

So, volume will increase $$16$$ times.

Final Step: Correct option (A).