States of Matter Test

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States of Matter Test - 31

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Weekly Quiz Competition

Question 1
1 / -0

An ideal solution has equal mol-fractions of two volatile components A and B in the vapour above the solution, the mol-fractions of A and B.

A
are both $$0.50$$

B
are equal but necessarily $$0.50$$

C
are not very likely to be equal

D
are $$1.00$$ and $$0.00$$ respectively

Solution
Question 2
1 / -0

Which of the following has maximum vapour pressure at a given temperature?

A

B

C

D

Solution

Vapour pressure increases with decreasing net molecular weight of a compound. Now among $$A,B,C$$ and $$D$$ in and $$D$$ intermolecular $$H-$$bonding is present as a result their net molecular weight increases but in $$A$$ and $$C$$ due to presence of intermolecular $$H-$$bonding their net molecular weight is approximately equals to their absolute molecular weight. Now intermolecular $$H-$$bonding is more feasible in $$C$$ due to presence of negative charge on $${ NO }_{ 3 }^{ \left( - \right) }$$ ion.
But absolute molecular weight of $$A$$ is less than $$C$$. So vapour pressure of $$A$$ will be highest.
Question 3
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Two liquids X and Y form an ideal solution. At $$300$$K, vapour pressure of the solution containing $$1$$ mol of X and $$3$$ mol of Y is $$550$$ mm Hg. At the same temperature, if $$1$$ mol of Y is further added to this solution, vapour pressure of the solution increases by $$10$$ mmHg. Vapour pressure (in mm Hg) of X and Y in their pure states will be ________ respectively.

A
$$200$$ and $$300$$

B
$$300$$ and $$400$$

C
$$400$$ and $$600$$

D
$$500$$ and $$600$$

Solution

Let
$$P=x_xP^0_x+x_yP^0_y$$

$$(\dfrac{1}{1+3})P_x^0+(\dfrac{3}{1+3})P_y^0=550\;mm\;of\;Hg$$

$$(\dfrac{1}{1+4})P_x^0+(\dfrac{4}{1+4})P_y^0=560\;mm\;of\;Hg$$

on solving for $$P_y^0$$ and $$P_x^0$$

We get,
$$P_x^0=400\;mm\;of\;Hg$$
$$P_y^0=600\;mm\;of\;Hg$$
Question 4
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If you are given Avogadro's number of atoms of a gas $$X$$. If half of the atoms are converted into $$X_{(g)}^+$$ by energy $$\Delta H$$. The IE of $$X$$ is :

A
$$\dfrac{2\Delta H}{N_A}$$

B
$$\dfrac{2N_A}{\Delta H}$$

C
$$\dfrac{\Delta H}{2N_A}$$

D
$$\dfrac{N_A}{\Delta H}$$

Solution

Given no. of atoms = Avogadro's no. of atoms = $${N}_{A} = 6.023 \times {10}^{23}$$
Given that $$\cfrac{{N}_{A}}{2}$$ atoms are ionized, i.e.,
Ionization energy of $$\cfrac{{N}_{A}}{2}$$ atoms of gas X = $$\Delta{H}$$
$$\therefore$$ Ionization energy of 1 atom of gas X = $$\cfrac{\Delta{H}}{\left( \cfrac{{N}_{A}}{2} \right)} = \cfrac{2. \Delta{H}}{{N}_{A}}$$
Hence, Ionisation energy of gas X is $$\cfrac{2. \Delta{H}}{{N}_{A}}$$.
Question 5
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What is the density of wet air with $$75\%$$ relative humidity at $$1$$ atm and $$300$$ K?
[Given: vapour pressure of $$H_2O$$ is $$30$$ torr and average molar mass of air is $$29$$g $$mol^{-1}$$]

A
$$1.614$$ g/L

B
$$0.96$$ g/L

C
$$1.06$$ g/L

D
$$1.164$$ g/L

Solution

Let us consider the equation

$${P_{wet\,air}} = \dfrac{{{P_d}}}{{{R_d}T}} + \dfrac{{{P_v}}}{{{R_v}T}} = \dfrac{{{P_d}{M_d} + {P_v}{M_v}}}{{RT}}$$

Given that :
$$T=$$ temperature $$=300\;K$$

$$M_d=$$ molar mass of dry air $$=29\;gmol^{-1}$$

$$M_v=$$ molar mass of water vapourr $$=18\;gmol^{-1}$$

$$R=0.821\;L\;atm\;mol^{-1}K^{-1} $$

$$P_v=30\;torr=0.0394\;atm$$

$$P_d=P-P_v=1-0.0394=0.9606\;atm$$

where $$P_d$$ partial pressure of dry air

$$R_d=$$ specific gas constant for dry air

$$R_v=$$ Specific gas constant for water vapour

$${P_{wet\,air}} = \dfrac{{0.9606 \times 29 + 0.0394 \times 18}}{{24.63}}$$

$$=1.164\;gL^{-1}$$

Hence, the correct option is $$D$$
Question 6
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Under $$3$$ atm, $$12.5$$ litre of a gas weighs $$15$$ gm. What will be the average speed of gaseous molecules?

A
$$8.028\times 10^4$$ cm/sec

B
$$6.028\times 10^2$$ cm/sec

C
$$8.028\times 10^5$$ cm/sec

D
$$6.028\times 10^4$$ cm/sec

Solution

Given,
Pressure = 3atm
Volume = 12.5 litre
Weight = 15 g
For gas We have $$PV = \dfrac{w}{m}RT$$
$$\implies 3×12.5=\dfrac{15}{m}×0.0821×T$$
$$T_m=30.45$$
Now,
$$ U_{AV}=\sqrt{\dfrac{8RT}{πm}}$$
$$=\sqrt{\dfrac{8×8.314×107×30.4×7}{22}}$$
$$=8.028×10^4cm\ sec^{−1}$$
Question 7
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A glass container is sealed with a gas at $$0.800$$ atm pressure and at $$25^o$$C. The glass container sustain pressure of $$2$$ atm. Calculate the temperature to which gas can be heated before bursting the container.

A
$$432^o$$C.

B
$$472^o$$C.

C
$$372^o$$C.

D
$$572^o$$C.

Solution

We have,

$$p_1=0.8$$atm$$,p_2=2$$atm and $$T_1=298$$k

Thus,

According to the Gas equation,

$$\dfrac{p_1}{T_1}=\dfrac{p_2}{T_2}$$

$$\dfrac{0.8}{298}=\dfrac{2}{T_2}$$

$$T_2=745K$$

$$T_2=472^0C$$

Hence, the correct option is $$\text{B}$$
Question 8
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A sample of water gas contains $$42\%$$ by volume of carbon monoxide. If the total pressure is $$760$$ mm of Hg, the partial pressure of carbon monoxide is :

A
$$380$$ mm of Hg

B
$$319.2$$ mm of Hg

C
$$38$$ mm of Hg

D
$$360$$ mm of Hg

Solution

$$1.$$ $$42\%$$ of CO is released
$$100\%$$ $$-$$ 28g
$$42\%$$ $$-$$ $$x$$

$$x = \dfrac{42\times28}{100}$$
$$x =11.76$$

$$2.$$ for $$ H_2O$$
100% $$-$$ 18g
58% $$-$$ $$x$$

$$x = \dfrac{58\times16}{100}$$

$$x = 10.44$$
mole of CO $$= \dfrac{11.76}{28} = 0.42$$

moles of $$H_2O$$ $$= \dfrac{10.44}{18} = 0.58$$

Dalton's law (p) $$= {P_T}\times{X_Q}$$

$$ = 760\times{\dfrac{0.42}{0.42+0.58}}$$

$$P = 319.2mm$$

The correct option is B.
Question 9
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If the pressure of $$N_2$$ and $$H_2$$ mixture in a closed apparatus is $$100$$ atm and $$20\%$$ of the mixture reacts then the pressure at the same temperature would be?

A
$$100$$

B
$$90$$

C
$$85$$

D
$$80$$

Solution

Solution:- (B) $$90$$

$${N}_{2} + 3 {H}_{2} \longrightarrow 2 N{H}_{3}$$

From the above reaction,

$$4$$ moles of reactant produces $$2$$ moles of $$N{H}_{3}$$

Therefore,

$$20 \%$$ mixture reacts to form $$10 \% \; N{H}_{3}$$.

Thus, $$80 \%$$ mixture and $$10 \% \; N{H}_{3}$$, i.e., $$90 \%$$ of mixture is left.

$$\because \; 100 \%$$ mixture has $$100 \; atm$$ pressure.

$$\therefore$$ Total pressure left $$= 90 \; atm$$
Question 10
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The vapour pressure of two liquids $$P$$ and $$Q$$ are $$80$$ torr and $$60$$ torr respectively. The total vapour pressure obtained by mixing $$3$$ mole of P and $$2$$ mole of $$Q$$ would be:

A
$$68$$ torr

B
$$20$$ torr

C
$$140$$ torr

D
$$72$$ torr

Solution

Mole fraction of $$P(X_P)=\cfrac {3}{3+2}={3}{5}$$
Mole fraction of $$Q(X_Q)=\cfrac {2}{3+2}=\cfrac {2}{5}$$
Now, Total vapour pressure $$(P_T)=X_PP_P+X_QP_Q$$
$$=\cfrac {3}{5}\times 80+\cfrac {2}{5}\times 60$$
$$\therefore$$ Total vapour pressure $$(P_T)=72 torr$$