States of Matter Test

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States of Matter Test - 37

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Weekly Quiz Competition

Question 1
1 / -0

At $${ 20 }^{ 0 }C$$, the vapour pressure of diethyl either is 44mm. When 6.4 g. of a non-volatile solute is dissolved in 50g. of either, the vapour pressure falls to 410mm. The Molecular weight of the solute is:

A
150

B
130

C
160

D
180

Solution

$$\Delta _{P}=440-410=30mm$$
$$\therefore \dfrac{\Delta _{P}}{P_{0}}=x_{solute}$$

$$\Rightarrow \displaystyle \frac{30}{440}=\frac{\frac{6.4}{M}}{\frac{6.4}{M}+\frac{50}{74}}$$

$$\Rightarrow \displaystyle \frac{19.2}{M}+\frac{150}{74}=\frac{44\times 6.4}{M}$$

$$\Rightarrow \displaystyle \frac{150}{74}=\frac{6.4(44-3)}{M}$$

$$\Rightarrow \displaystyle M=\frac{6.4\times 41\times 74}{150}$$

$$\approx 130$$
Question 2
1 / -0

Equal masses of $${ SO }_{ 2, }{ CH }_{ 4 }$$ and $${ CH }_{ 4 }$$ are mixed in empty container at 298 K, when total pressure is 2.1 atm. The partial pressure of $$O_2$$ in the mixture is:

A
$$0.5$$ atm

B
$$0.75$$ atm

C
$$1.2$$ atm

D
$$0.6$$ atm
Question 3
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18 g glucose is added to 178.2 g of water the vapour pressure of water for this aqueous solution at $${ 100 }^{ 0 }C$$ is:

A
704 torr

B
759 torr

C
7.6 torr

D
None of these

Solution

Molar mass of water $$=18$$ $$g$$

For $$178.2$$ $$g$$ of $$H_2$$ ,$$n_A = 9.9$$ $$X_B= \cfrac {0.1}{0.1 + 9.9} =0.01$$

Molar mass of Glucuse $$=180$$ $$g$$ $$X_A = 0.99$$

For $$18$$ $$g$$ , $$n_B = 0.1$$

For lowering of vapour pressure ,

$$P = P^o \times A= P^oA(1-X_B) = 760(1-0.01)$$

$$P = 760-7.6$$

$$P= 752.4$$ $$torr $$
Question 4
1 / -0

Which of the following curve does not represent Gay-lussac's law?

A

B

C

D

Solution

Gay-lussac's law.
So,the correct answer is B
Question 5
1 / -0

Which of the liquids in each of the following pairs has a higher vapour pressure?

A
Alcohol, glycerine

B
Mercury, water

C
petrol, kerosene

D
None of these

Solution

Alcohol and glycerine have high vapour pressure because they have high intermolecular attraction.
As the intermolecular attraction increases vapour pressure increases.
Question 6
1 / -0

Vapour pressure of $${ CCI }_{ 4 }$$ at $${ 24 }^{ o }C$$ is $$143 mmHg 0.05$$g of the non-volatile solute (mol.wt.=65) is dissolved in $$100 ml { CCI }_{ 4 }$$. Find the vapour pressure of the solution (Density of $${ CCI }_{ 4 }$$=$$1.58 g/{ cm }^{ 2 }$$)

A
$$143.99 mm$$

B
$$94.39 mm$$

C
$$199.34 mm$$

D
$$14.197 mm$$

Solution

The equation of relative lowering vapour pressure is given as-
$$\cfrac{{p}^{0} - p}{{p}^{0}} = {X}_{2}$$
Whereas,
$${p}^{0} =$$ Vapour prssure of solvent
$$p =$$ Vapour pressure of solution
$${X}_{2} =$$ Mole fraction of solute
Given that density of $$C{Cl}_{4}$$ is $$1.58 \; {g}/{{cm}^{3}}$$
As we know that,
$$\text{density} = \cfrac{\text{mass}}{\text{volume}}$$
$$\Rightarrow 1.58 = \cfrac{\text{mass}}{100}$$
$$\Rightarrow \text{mass of } C{Cl}_{4} = 158 \; g$$
Molecular weight of $$C{Cl}_{4} = 154 \; g$$
$$\therefore$$ No. of moles of $$C{Cl}_{4} = \cfrac{158}{154} = 1.026 \text{ mol}$$
Given weight of solute $$= 0.05 \; g$$
Molecular weight of solute $$= 65 \; g$$
$$\therefore$$ No. of moles of solute $$= \cfrac{0.05}{65} = 0.00077 \text{ mol}$$
$$\therefore$$ Total no. of moles $$= 1.026 + 0.00077 = 1.02677 \text{ mol}$$
$$\therefore$$ Mole fraction of solute $$= \cfrac{0.00077}{1.02677} = 0.00075$$
Vapour pressure of solvent $$= 143 \text{ mm of Hg}$$
Noe from equation of relative lowering vapour pressure, we have
$$\cfrac{143 - p}{143} = 0.00075$$
$$\Rightarrow 143 - p = 0.107$$
$$\Rightarrow p = 142.893 \text{ mm of Hg}$$
Hence the vapour pressure of solution is $$142.893 \text{ mm of Hg}$$.
Question 7
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The vapour pressure of water at $$20$$ is 17.5 mm Hg. If 18 g of glucose $${ C }_{ 6 }{ H }_{ 12 }{ O }_{ 6 }$$ is added to 178.2 g water $$20$$ , the vapour pressure of the resulting solution will be:

A
17.675 mm Hg

B
15.750 mm Hg

C
16.500 mm Hg

D
17.325 mm Hg

Solution

No. of moles of glucose=$$\cfrac{18}{180}=0.1$$

No. of moles of water=$$\cfrac{178.2}{18}=9.9$$

Mole fraction of glucose=$$\cfrac{0.1}{4.9+0.1}=\cfrac{0.1}{10}=0.01$$

Using Raoult's law, as solute is non-volatile,

$${ x }_{ \beta }=\cfrac { { P }_{ A }^{ o }-{ P }_{ A } }{ { P }_{ A }^{ o } } $$

$$P_{A}^{o}=$$vapour pressure of pure water

$$P_{A}=$$vapour pressure of the solution

$$x_{\beta}=$$mole fraction of glucose

$$\Longrightarrow 0.01=\cfrac { 17.5-{ P }_{ A } }{ 17.5 } \\ \Longrightarrow { P }_{ A }=17.5-0.175=17.325$$

The vapor pressure of the mixture=17.325 mm Hg.

Hence, the correct option is $$D$$
Question 8
1 / -0

The vapour pressure of o-nitrophenol at any given temperature is predicted to be:

A
Higher than that of p-nitrophenol

B
Lower than that of p-nitrophenol

C
Same as that of p-nitrophenol

D
Higher or lower depending upon the size of the vessel

Solution

In o- nitrophenol , there is intra-molecular as as well as inter-molecular hydrogen bonding while in p-nitrophenol there is only inter-molecular hydrogen bonding. Therefore, it is difficult to vaporize o-nitrophenol than p-nitrophenol . Hence the vapour pressure of o-nitrophenol is higher than that of p-nitrophenol.
Question 9
1 / -0

Equal weights of $$HF$$, $$HCl$$ ,$$HBr$$ , and $$HI$$ $${at\ 87}^{0}C$$ and $$750\ mm$$ $$Hg$$ pressure are taken. The correct sequence of these gases in the increasing order of their volume is ?

A
HI<HBr<HCl<HF

B
HBr<HI<HCl<HF

C
HF<HBr<HCl<HI

D
HI<HBr<HF<HCl

Solution

At given temperature and pressure, volume of a gas depends on its number of moles. So volume of a gas is directly proportional to the number moles of gas i.e. mass of the gas divided by molecular mass of that gas. Since all gases have equal weights, the gas having minimum molecular weight will have maximum volume. Now here increasing order of molecular weight is :
HF<HCl<HBr<HI
therefore increasing order of their volume will be : HI<HBr<HCl<HF
Question 10
1 / -0

An open vessel at $$300\ K$$ is heated till $$\dfrac { 2 }{ 5 } $$ of the air in it expelled. Assuming that the volume of the vessel remains constant, the temperature to which the vessel is heated, is

A
$$750 K$$

B
$$400 K$$

C
$$500 K$$

D
$$1500 K$$

Solution

We know at constant V and P:

$$ n_{1} T_{1} = n_{2} T_{2} \hspace{20mm} \text{[ $\because P_{1} V_{1}$ constant ]} $$

$$n_1 = n$$ and $$n_2 = n-\dfrac{2n}{5}$$ and $$T_1 = 300K$$

$$n(300) = (n-\cfrac{2}{5}n) (T_{2}) = (\cfrac{3n}{5}) T_{2} $$

$$T_{2} = 500 K$$