States of Matter Test

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States of Matter Test - 27

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Weekly Quiz Competition

Question 1
1 / -0

If the concentration of water vapour in the air is $$1\%$$ and the total atmospheric pressure equals $$1$$atm then the partial pressure of water vapour is?

A
$$0.1$$ atm

B
$$1$$ mm Hg

C
$$7.6$$ mm Hg

D
$$100$$ atm

Solution

According to Dalton's law of partial pressures,
pressure $$\propto \Lambda_{\text{water vapour}}$$
Pressure of water vapour=$$\left( \cfrac { 1 }{ 100 } \right) 1$$ atm=0.01 atm
1 atm$$\Longrightarrow $$ 760 mm of Hg
0.01 atm$$\Longrightarrow $$ 7.6 mm of Hg
Question 2
1 / -0

Two moles of pure liquid 'A' ($$P^o_A$$ = 80mm of Hg) and 3 moles of pure liquid 'B' ($$P^o_B$$ = $$120$$mm of $$Hg$$) are mixed. Assuming ideal behaviour.

A
Vapour pressure of the mixture is $$104$$ mm of $$Hg$$

B
Mole fraction of liquid '$$A'$$ in Vapour pressure is $$0.3077$$

C
Mole fraction of '$$B$$' in vapour pressure is $$0.692$$

D
Mole fraction of '$$B$$' in vapour pressure is $$0.785$$

Solution

Formula of Ammonia = $$NH_{3}$$

$$P_{A}^{\circ}=80 mm Hg$$ $$n_{A}=2$$ mole

$$P_{B}^{\circ}=120 mm Hg$$ $$n_{B}=3$$ mole

$$x_{A}=\frac{n_{A}}{n_{A}+n_{B}}=\frac{2}{2+3}=0.4$$

$$x_{B}=\frac{n_{B}}{n_{A}+n_{B}}=\frac{3}{2+3}=0.6$$

Vapour pressure $$P_{A}=P_{A}^{\circ}x_{A}+P_{B}^{\circ}x_{B}$$

$$P_{A}=80\times 0.4+120\times 0.6$$

$$=104 mmHg$$

Option A is correct
Question 3
1 / -0

Two flasks of equal volume have been joined by a narrow tube of negligible volume. Initially, both flasks are at 300K containing 0.6 mole of $$O_2$$ at 0.5 atm pressure. one of the flasks is placed in the thermostat of 600K. FInd the number of mole of $$O_2$$ gas in each flask.

A
0.5mol and 0.3 mol

B
0.5mol and 0.2 mol

C
0.4mol and 0.2 mol

D
0.6mol and 0.3 mol

Solution

$$ 0.5 V_1 =0.6R\times 300$$
$$ V = 360R$$
in 2nd case the pressure will be same for both flask and sum of moles is 0.6
Flask-1
$$ P\times 360R = n_1R\times 300$$
$$ n_1 =1.2P$$
Flask-2
$$ P\times 360R = n_2R\times 600$$
$$ n_2 = 0.6P$$
$$ n_1+n_2 = 0.6$$
$$ 1.2P+0.6P =0.6$$
$$ 1.8P = 0.6$$
Final pressure-
$$ P = \dfrac{1}{3}$$
Number of moles of oxygen in flasks-
$$n_1 = 1.2\times \dfrac{1}{3}= 0.4$$
$$n_2 = 0.6\times \dfrac{1}{3} = 0.2$$
Question 4
1 / -0

Which of the following relationships for various gas laws is not correct?

A
$$V_{t} = V_{0} + \dfrac {V_{0}}{273}\times t$$

B
$$\dfrac {V_{1}}{T_{1}} = \dfrac {V_{2}}{T_{1}} (constant\ P)$$

C
$$\dfrac {P_{1}}{T_{1}} = \dfrac {P_{2}}{T_{1}} (constant\ V)$$

D
$$\dfrac {P_{1}T_{1}}{V_{1}} = \dfrac {P_{2}T_{2}}{V_{2}}$$

Solution

According to ideal gas equation $$PV=nRT$$
$$\therefore { P }_{ 1 }{ V }_{ 1 }=nR{ T }_{ 1 }$$ or $${ P }_{ 1 }{ V }_{ 1 }/{ T }_{ 1 }=nR$$
$${ P }_{ 2 }{ V }_{ 2 }=nR{ T }_{ 2 }$$ or $${ P }_{ 2 }{ V }_{ 2 }/{ T }_{ 2 }=nR$$
$$\Rightarrow \cfrac { { P }_{ 1 }{ V }_{ 1 } }{ { T }_{ 1 } } =\cfrac { { P }_{ 2 }{ V }_{ 2 } }{ { T }_{ 2 } } $$
Hence $$\cfrac { { P }_{ 1 }{ T }_{ 1 } }{ { V }_{ 1 } } =\cfrac { { P }_{ 2 }{ T }_{ 2 } }{ { V }_{ 2 } } $$ is incorrect
Question 5
1 / -0

Which of the following values does not represent the correct value of $$R$$?

A
$$8.314\ Pa\ m^{2} K^{-1} mol^{-1}$$

B
$$8.314\times 10^{-2} bar\ L\ K^{-1} mol^{-1}$$

C
$$0.0821\ J\ K^{-1} mol^{-1}$$

D
$$8.314\ J\ K^{-1} mol^{-1}$$

Solution

The value of ideal gas constant in vaious units can be given as follow:
$$R = 8.314\ Pa\ m^{3}\ K^{-1} mol^{-1}$$
$$R = 8.314\times 10^{-2} bar\ L\ K^{-1} mol^{-1}$$
$$R = 8.314\ J\ K^{-1} mol^{-1}$$
$$R = 0.0821\ L\ atm\ K^{-1} mol^{-1}$$.
Thus value given in option C is incorrect.
Question 6
1 / -0

How many number of moles of nitrogen will be present in $$2.24\ L$$ of nitrogen gas at $$STP?$$

A
$$9.9$$

B
$$0.099$$

C
$$0.001$$

D
$$1.00$$

Solution

$$PV = nRT$$

$$P = 1\ atm, T = 273\ K, V = 2.24\ L$$

$$R = 0.0821\ L\ atm\ K^{-1} mol^{-1}$$

$$n = \dfrac {1\times 2.24}{0.0821\times 273} = 0.099$$ moles.
Question 7
1 / -0

The statements for laws of chemical combinations are given below. Mark the statement which is not correct.

A
Matter can neither be created nor destroyed: Law of conservation of mass

B
A compound always contains exactly the same proportion of elements by weight: Law of definite proportions

C
When gases combine they do so in a simple ratio by weight: Gay Lussac's Law

D
Equal volumes of gases at same temperature and pressure contain same number of molecules: Avogadro's Law

Solution

Law of Conservation of Mass states that matter can neither be created nor destroyed. The total mass, that is, the sum of the mass of the reacting mixture and the products formed remains constant.

Law of Definite Proportions states that the proportion of elements by weight in a given compound will always remain exactly the same.

Gay Lussac’s Law of Gaseous Volume states that in a gaseous system, when the reactants combine together they always do so in volumes which bear a simple ratio to one another and to the volume of the products, under similar conditions of temperature and pressure.

Avogadro’s Law states that under the same conditions of temperature and pressure, the equal volume of all the gases contains an equal number of molecules.

Thus statement C is for Law of definite proportion.
Option C is correct.
Question 8
1 / -0

What volume in litres will be occupied by $$4.4$$ g of $$CO_2$$ at STP?

A
$$22.4\ L$$

B
$$44.8\ L$$

C
$$12.2\ L$$

D
$$2.24\ L$$

Solution

$$44\ g$$ of $$CO_{2}$$ occupies $$22.4\ L$$
$$4.4\ g$$ of $$CO_{2}$$ occupies $$\dfrac {22.4}{44}\times 4.4 = 2.24\ L$$.
Question 9
1 / -0

$$6 g$$. of the area of dissolved in $$90 g$$. of boiling water. The vapour pressure of the solution is:

A
$$744.8$$ mm

B
$$758$$ mm

C
$$761$$ mm

D
$$760$$ mm

Solution

Vapour pressure $$P=P°x,\quad P°=$$Boiling point of pure solvent
$$P=P°\cfrac { { n }_{ 1 } }{ { n }_{ 1 }+{ n }_{ 2 } }$$

$$=760\left( \cfrac { 0.1 }{ 0.1+5 } \right) \quad \quad 90g{ H }_{ 2 }O=5mol={ n }_{ 2 };\quad 6g$$ urea$$=0.1mol={ n }_{ 1 }$$

$$=760\left( \cfrac { 0.1 }{ 0.1+5 } \right) $$

$$=760\times 0.0196=14.89 $$ Torr=lowered pressure

Now, 1 torr = 1 mm of Hg

Vapour pressure of solution$$=760-14.89\approx 744.8 mm\ of\ Hg$$

Option A is correct.
Question 10
1 / -0

Which of the following statement about Avogadro's hypothesis is correct?

A
Under similar conditions of temperature and pressure, gases react with each other in simple ratio.

B
Under similar conditions of temperature and pressure, equal volumes of all gases contain same number of molecules

C
At NTP all gases contain same number of molecules

D
Gases always react with gases only at the given temperature and pressure

Solution

Avogadro’s Law states that under same conditions of temperature and pressure, equal volume of all the gases contain equal number of molecules.

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States of Matter Test - 27

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States of Matter Test - 27

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Question 1

$$0.1$$ atm

$$1$$ mm Hg

$$7.6$$ mm Hg

$$100$$ atm

Solution

Question 2

Vapour pressure of the mixture is $$104$$ mm of $$Hg$$

Mole fraction of liquid '$$A'$$ in Vapour pressure is $$0.3077$$

Mole fraction of '$$B$$' in vapour pressure is $$0.692$$

Mole fraction of '$$B$$' in vapour pressure is $$0.785$$

Solution

Question 3

0.5mol and 0.3 mol

0.5mol and 0.2 mol

0.4mol and 0.2 mol

0.6mol and 0.3 mol

Solution

Question 4

$$V_{t} = V_{0} + \dfrac {V_{0}}{273}\times t$$

$$\dfrac {V_{1}}{T_{1}} = \dfrac {V_{2}}{T_{1}} (constant\ P)$$

$$\dfrac {P_{1}}{T_{1}} = \dfrac {P_{2}}{T_{1}} (constant\ V)$$

$$\dfrac {P_{1}T_{1}}{V_{1}} = \dfrac {P_{2}T_{2}}{V_{2}}$$

Solution

Question 5

$$8.314\ Pa\ m^{2} K^{-1} mol^{-1}$$

$$8.314\times 10^{-2} bar\ L\ K^{-1} mol^{-1}$$

$$0.0821\ J\ K^{-1} mol^{-1}$$

$$8.314\ J\ K^{-1} mol^{-1}$$

Solution

Question 6

$$9.9$$

$$0.099$$

$$0.001$$

$$1.00$$

Solution

Question 7

Matter can neither be created nor destroyed: Law of conservation of mass

A compound always contains exactly the same proportion of elements by weight: Law of definite proportions

When gases combine they do so in a simple ratio by weight: Gay Lussac's Law

Equal volumes of gases at same temperature and pressure contain same number of molecules: Avogadro's Law

Solution

Question 8

$$22.4\ L$$

$$44.8\ L$$

$$12.2\ L$$

$$2.24\ L$$

Solution

Question 9

$$744.8$$ mm

$$758$$ mm

$$761$$ mm

$$760$$ mm

Solution

Question 10

Under similar conditions of temperature and pressure, gases react with each other in simple ratio.

Under similar conditions of temperature and pressure, equal volumes of all gases contain same number of molecules

At NTP all gases contain same number of molecules

Gases always react with gases only at the given temperature and pressure

Solution

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