States of Matter Test

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States of Matter Test - 65

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Weekly Quiz Competition

Question 1
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The empirical formula of a monobasic acid is $${CH}_{2}O$$. The vapour density of its ethyl ester is $$44$$. What is the Molecular formula of the acid?

A
$${C}_{2}{H}_{4}{O}_{2}$$

B
$${CH}_{2}{O}_{2}$$

C
$${C}_{2}{H}_{2}{O}_{2}$$

D
$${C}_{3}{H}_{6}{O}_{3}$$

Solution

Empirical formula $$CH_{2}O\,\,\rightarrow weight=12+2+16=30\,gm$$
Molecular mass of acid=(Molecular mass of ethyl ester)-(Molecular weight of $$C_{2}H_{5}$$)+(Atomic mass of $$H$$ atom)
$$=(2\times V\cdot D)-(2\times 12\,+5\times 1)+1$$
$$=88-29+1=60$$
$$\therefore$$ Molecular formula$$=\cfrac{Molecular \,Weight}{Empirical \,Weight}\times Empirical \,Formula$$
$$=\cfrac{60}{30}\times C(H_{2}O)=C_{2}H_{4}O_{2}$$
Question 2
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Directions For Questions

When a vapour consisting of a mixture of colourless gas X and a brown gas Y, at atmospheric pressure is gradually heated from $$25^oC$$, its colour is found to deepen at first and then to fade as the temperature is raised above $$160^oC$$. At $$600^oC$$, the vapour is almost colourless consisting of a gas Z and $$O_2$$ but it's colour deepens when the pressure is raised at this temperature.

...view full instructions

The colourless gas X in the mixture is:

A
Bromine

B
Pure nitrogen dioxide

C
Dinitrogen tetraoxide

D
Nitric oxide

Solution

Given,
Colorless gas - $$X$$
Brown gas - $$Y$$
The bromine does not change its color at a given temperature, but nitrogen dioxide and dinitrogen tetraoxide change their color at a given temperature and give the same color as given in the question.
heat dinitrogen tetraoxide
$${N_2}{O_4}$$ $$\xrightarrow{{{{160}^{\rm O}}C}}$$ $$2N{O_2}$$ $$\xrightarrow{{{{600}^{\rm O}}C}}$$ $$2NO$$ $$ + $$ $${O_2}$$
in this reaction,
$${N_2}{O_4}$$ - dinitrogen tetraoxide (colorless)
$$2N{O_2}$$ - nitrogen dioxide (brown)
$$2NO$$ - Nitric oxide (colorless)
Therefore, the name of $$X$$gas is nitric oxide that is colorless.

Hence, option(d) is the correct answer.
Question 3
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Plots of $$\cfrac { PV }{ RT } vsP$$ for 1 mole of $${ H }_{ 2 }$$, $${ NH }_{ 3 }$$ and $${ CH }_{ 4 }$$ gases are given. Match the curve with corresponding gases
Curve Gas
(i) Curve 'a' p- $${ H }_{ 2 }$$
(ii) Curve 'b' q- $${ NH }_{ 3 }$$
(iii) Curve 'c' r- $${ CH }_{ 4 }$$

A
(i)-p, (ii)-q, (iii)-r

B
(i)-q, (ii)-r, (iii)-p

C
(i)-p, (ii)-r, (iii)-q

D
(i)-r, (ii)-q, (iii)-p

Solution
Question 4
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Moles of $$Na_2SO_4$$ to be dissolved in 12 mole of water to lower its vapour pressure by 10 millimeter Mercury at a temperature at which vapour pressure of pure water is 50 millimeters?

A
1.5 moles

B
2 moles

C
1 mole

D
3 moles
Question 5
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According to kinetic theory of gases :

A
collisions are always elastic

B
between collisions, the molecules move in straight lines with constant velocities

C
only a small number of molecules have a very high velocity

D
at of the above
Question 6
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3.6 gm of $$O_2$$ is adsorbed on 1.2 gr of metal powder. What volume of $$O_2$$ adsorbed per gram of the absorbant at 1 atm and 273K?

A
2.1

B
0.19

C
1

D
None

Solution

$$1.2 g \rightarrow 3.6 g$$ oxygen
$$1 g \rightarrow \dfrac{3.6}{1.2} = 3 g$$
$$PV = \dfrac{x}{M} \times RT$$
$$1 \times V = \dfrac{3}{32} \times 0.0821 \times 273$$
$$V=2.1\ L$$
Question 7
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The pressure exerted by a mass of x mg resting on the area of 1.00 $${ cm }^{ 2 }$$ is 1.00 Pa, then x is

A
10.6

B
10.3

C
103

D
10.2

Solution
Question 8
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An inflated balloon has a volume of $$6.0L$$ at sea level ($$P=1atm$$) and is allowed to ascend in altitude until the pressure is $$0.4atm$$. During the rise, the temperature of the gas falls from $${27}^{o}C$$ to $$-{23}^{o}C$$. The volume of balloon at its final altitude is

A
$$12.5L$$

B
$$15L$$

C
$$10L$$

D
$$3L$$

Solution
Question 9
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The solubility of a specific non-volatile salt is 4g in 100 g of water at $${25}^{0}$$C. If 2.0g, 4.0g and 6.0g of the salt added of 100g of water at $${25}^{0}$$C, in system X, Y and Z. The vapour pressure would be in the order:

A
$$Z>Y>X$$

B
$$X>Y>Z$$

C
$$Z>X=Y$$

D
$$X>Y=Z$$

Solution
Question 10
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The vapour pressure of pure water at $$25^{\circ}C$$ is $$30\ mm$$. The vapour pressure of $$10\%\ (W/W)$$ glucose solution at $$25^{\circ}C$$ is :

A
$$31.5\ mm$$

B
$$30.6\ mm$$

C
$$29.67\ mm$$

D
$$26.56\ mm$$

Solution

$$10$$% solution $$\Rightarrow 10g$$ of glucose , $$90g$$ of water in $$100gm$$ of solution

So, number of moles of water in $$90gm$$=$$ \cfrac {90}{18}= 5$$ moles

Number of moles of glucose= $$\cfrac {10}{180}=0.056$$ moles

So, total moles= $$5+0.056=5.056$$ moles

Mole fraction of water= $$5/5.056=0.988$$

So, final vapour pressure= $$P\times X_{water}=30\times 0.980=29.6\ mm$$

Hence, the correct option is $$C$$