States of Matter Test

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States of Matter Test - 41

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Weekly Quiz Competition

Question 1
1 / -0

The molecular weight of a gas is 40. At 400K, if 120g of this gas has a volume of 20 litres, the pressure of the gas is?

A
5.02 atm

B
0.546 atm

C
4.92 atm

D
49.6 atm

Solution

Moles of gas $$=\cfrac{120}{40}=3=n$$
$$\because PV=nRT$$
$$\therefore P\times 20=3\times 0.0821\times 400$$
$$P=4.926$$ atm.
Question 2
1 / -0

An open flask has Helium gas at 2atm and 327$$^{\circ}C$$. The flask is heated at 527$$^{\circ}C$$ at the same pressure. The fraction of original gas remaining in the flask is?

A
$$\frac{3}{4}$$

B
$$\frac{1}{4}$$

C
$$\frac{1}{2}$$

D
$$\frac{2}{5}$$

Solution

Initial pressure, $$P_1= 1 atm$$
Initial temperature, $$T_1= 327^oC=600K$$
Final pressure, $$P_2= 1 atm$$
Final temperature, $$T_2= 527^oC= 800K$$
now, $$\cfrac {P_1V_1}{T_1}=\cfrac {P_2V_2}{T_2}$$
$$\Rightarrow V_2=\cfrac {1 \times V_1}{600}\times \cfrac {800}{1}=\cfrac {4}{3}V_1$$
But the flask can hold only $$V_1$$ volume of air,
$$\therefore$$ Volume of air expelled= $$\cfrac {4}{3}V_1-V_1=\cfrac {V_1}{3}$$
$$\therefore$$ Fraction of air expelled out= $$\left(\cfrac {V_1}{3}\right)/(4V_1/3)$$
$$\therefore$$ Fraction of air remaining in flask= $$1-\cfrac {1}{4}=3/4$$
Question 3
1 / -0

At $$127^0C$$ and 1 atm pressure, a mixture of a gas contains 0.3 mole of $$N_2$$ and 0.2 mole of $$O_2$$. The volume of the mixture is?

A
15 L

B
18.2 L

C
16.4 L

D
22.4 L

Solution

Let $$V$$ be volume of mixture.
Partial pressure of $$O_2=X_{O_2}\times P_{total}=\cfrac{n_{O_2}RT}{V}$$
$$\therefore V=\cfrac{n_{O_2}RT}{X_{O_2}\times P_{total}}=\cfrac{0.2\times 0.0821\times 400}{\left(\cfrac{0.2}{0.5}\right)\times 1}=16.4L$$
Question 4
1 / -0

Dry ice is solid carbon dioxide. A $$0.05\ g$$ sample of dry ice is placed in an evacuated $$4.6$$L vessel at $$30^0$$C. Calculate the pressure inside the vessel after all the dry ice has been converted to $$CO_2$$ gas.

A
$$6.14$$ atm

B
$$0.614$$ atm

C
$$0.0614$$ atm

D
$$6.14 \times 10^{-3}$$ atm

Solution
Question 5
1 / -0

A mixture of ethanol and propanol has vapor pressure 279 mm of Hg at 300 K. The mol fraction of ethanol in the solution is 0.6 and vapour pressure of pure propanol is 210 mm of Hg. Vapour pressure of pure ethanol will be:

A
$$336.6$$ mm

B
$$325$$ mm

C
$$375$$ mm

D
$$415$$ mm

Solution

$$p=p_1+p_2$$
$$\Rightarrow p={p_1}^0x_1+{p_2}^0x_2$$ ( Raoult's law )
$$p={p_1}^0\left( 1-x_2\right) +{p_2}^0x_2$$
$$p={p_1}^0+\left( {p_2}^0-{p_1}^0\right) x_2$$
$$279mm={p_1}^0+\left( {p_2}^0-{p_1}^0\right) x_2$$
Let $$2,$$ ethanol
$$1,$$ propanol
$$\therefore 279mm=210mm+\left( {p_2}^0-210\right) 0.6$$
$$69=\left( {p_2}^0-210\right)0.6$$
$${p_2}^0-210=115$$
$$\Rightarrow {p_2}^0=325mm$$
Hence, the answer is $$325mm.$$
Question 6
1 / -0

A gas at a pressure of $$5\ atm$$ is heated from $$0^{o}$$ to $$546^{o}C$$ and simultaneous compressed to $$\dfrac {1}{3}rd$$ of it original volume. Hence final pressure is?

A
$$10\ atm$$

B
$$45\ atm$$

C
$$30`\ atm$$

D
$$5\ atm$$

Solution

Here $$\cfrac{PV}{T}=constant$$
Given $$V_2=\cfrac{V_1}{3},$$ $$P_1=5\text{ }atm, \\T_1=273K,\\T_2=819K$$
$$\therefore \cfrac{P_1V_1}{T_1}=\cfrac{P_2V_2}{T_2}$$
$$\therefore P_2=P_1(\cfrac{V_1}{V_2})(\cfrac{T_2}{T_1})=5\times 3\times \cfrac{819}{273}=45\text{ }atm$$
Question 7
1 / -0

The vapour pressure of a pure liquid A is 70 torr at 300 K. It forms an ideal solution with another liquid B. The mole fraction of B in the solution is 0.2 and total pressure of solution is 84 torr at 300 K. The vapour pressure of pure liquid B at 27C is ?

A
0.14 torr

B
560 torr

C
140 torr

D
70 torr

Solution

Solution:- (C) $$140 \text{ torr.}$$
Mole fraction of $$B = 0.2$$

As we know that,
$${x}_{A} + {x}_{B} = 1$$
$$\Rightarrow {x}_{A} = 1 - 0.2 = 0.8$$
$${{p}^{°}}_{A} = 70 \text{ torr.}$$
$${{p}^{°}}_{B} = ?$$
$${p}_{total} = 84 \text{ torr.}$$

From Dalton's law of partial pressure,
$${p}_{total} = {p}_{A} + {p}_{B} ..... \left( 1 \right)$$

From Raoult's law of vapour pressure,
$${p}_{A} = {{p}^{°}}_{A} \cdot {x}_{A}$$
$$\Rightarrow {p}_{A} = 70 \times 0.8 = 56$$

Again, from Raoult's law of vapour pressure,
$${p}_{B} = {{p}^{°}}_{B} \cdot {x}_{B}$$
$${p}_{B} = {{p}^{°}}_{B} \times 0.2$$

Substituting these values in equation $$\left( 1 \right)$$, we have
$$84 = 56 + 0.2 \times {{p}^{°}}_{B}$$
$$\Rightarrow {{p}^{°}}_{B} = \cfrac{84 - 56}{0.2} = 140 \text{ torr.}$$

Hence the vapour pressure of pure liquid $$B$$ is $$140 \text{ torr.}$$.
Question 8
1 / -0

The boiling point of $$ C_{6}H_{6},CH_{3}OH, C_{6}H_{5}NH_{2} $$ and $$C_{6}H_{5}NO_{2} $$ are $$ 80^{\circ},\ 65 ^{\circ},\ 184^{\circ}$$ and $$212^{\circ} $$ respectively. Which will show highest vapourr pressure at room temperature?

A
$$ C_{6}H_{6}$$

B
$$ CH_{3}OH$$

C
$$ C_{6}H_{5}NH_{2} $$

D
$$C_{6}H_{5}NO_{2} $$

Solution

Boiling point is the temperature at which the vapor pressure of a liquid become equal to atmospheric pressure.
$$\rightarrow$$ So, the compounds which has low boiling point will have high vapor pressure.
$$\rightarrow$$ Given $${ CH }_{ 3 }OH$$ has boiling point of $${ 65 }^{ 0 }$$.
$$\rightarrow$$ It has highest vapor among given with $$0.166$$ atm.
Ans :- Option B.
Question 9
1 / -0

Kinetic energy of gas depends on:

A
temperature

B
pressure

C
surface area

D
both $$A$$ and $$B$$

Solution

We know that $$K.E=\dfrac { 3 }{ 2 } nRT$$
Here $$T$$ denotes temperature and this energy doesn't depend on any other parameter.
Ans :- Option A.
Question 10
1 / -0

One mole of non volatile solute is dissolved in two moles of water. The vapour pressure of the solution relative to that of water is:

A
$$\dfrac{2}{3}$$

B
$$\dfrac{1}{3}$$

C
$$\dfrac{1}{2}$$

D
$$\dfrac{3}{1}$$

Solution

Relative lowering in vapor pressure $$=\cfrac{P^{\circ}-P_s}{P^{\circ}}=x_A$$
Where $$x_A=$$ mole fraction of non-volatile solute.
$$\therefore \cfrac{P^{\circ}-P_s}{P^{\circ}}=\cfrac{1}{3}\\ \therefore \cfrac{P_s}{P^{\circ}}=1-\cfrac{1}{3}=\cfrac{2}{3}$$

Hence, the correct option is $$A$$

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States of Matter Test - 41

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States of Matter Test - 41

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SHARING IS CARING

Question 1

5.02 atm

0.546 atm

4.92 atm

49.6 atm

Solution

Question 2

$$\frac{3}{4}$$

$$\frac{1}{4}$$

$$\frac{1}{2}$$

$$\frac{2}{5}$$

Solution

Question 3

15 L

18.2 L

16.4 L

22.4 L

Solution

Question 4

$$6.14$$ atm

$$0.614$$ atm

$$0.0614$$ atm

$$6.14 \times 10^{-3}$$ atm

Solution

Question 5

$$336.6$$ mm

$$325$$ mm

$$375$$ mm

$$415$$ mm

Solution

Question 6

$$10\ atm$$

$$45\ atm$$

$$30`\ atm$$

$$5\ atm$$

Solution

Question 7

0.14 torr

560 torr

140 torr

70 torr

Solution

Question 8

$$ C_{6}H_{6}$$

$$ CH_{3}OH$$

$$ C_{6}H_{5}NH_{2} $$

$$C_{6}H_{5}NO_{2} $$

Solution

Question 9

temperature

pressure

surface area

both $$A$$ and $$B$$

Solution

Question 10

$$\dfrac{2}{3}$$

$$\dfrac{1}{3}$$

$$\dfrac{1}{2}$$

$$\dfrac{3}{1}$$

Solution

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