States Of Matter(12th Grade > Chemistry ) Questions and answers for exam Preparation

12th Grade > Chemistry

STATES OF MATTER MCQs

Total Questions : 30 | Page 2 of 3 pages

Question 11. The temperature at which RMS velocity of

S O_{2}

molecules is half that of He molecules at 300 K is

150 K
600 K
900 K
1200 K

Discuss Question

Answer: Option D. -> 1200 K
:
D

\frac{U_{S O_{2}}}{U_{H} e} = \frac{1}{2} = \sqrt{\frac{M_{H e}}{M_{S O_{2}}} \frac{T_{S O_{2}}}{T_{H e}}} = \sqrt{\frac{4}{64} \frac{T_{S O_{2}}}{300}}

= \frac{4}{64} \frac{T_{S O_{2}}}{300} = \frac{1}{4}; T_{S O_{2}} = 1200^{\circ} K

Question 12. The temperature and pressure at which ice, liquid water and water vapour can exist together are

0∘ C, 1atm
2∘C, 4.7atm
0∘C, 4.7mm
2∘C, 4.7mm

Discuss Question

Answer: Option C. -> 0∘C, 4.7mm
:
C
All the three phases of water can coexist at 0

^{\circ}

C & 4.7 mm pressure.

Question 13.

n_{1}

and

n_{2}

moles of two ideal gases (mol. wt.

M_{1}

and

M_{2}

) respectively at temperature

T_{1}

K and

T_{2}

K are mixed. Assuming that no loss of energy, the temperature of mixture becomes -

n1T1+n2T2
n1T1+n2T2T1+T2
n1T1+n2T2n1+n2
n1×T2n1×n2

Discuss Question

Answer: Option C. -> n1T1+n2T2n1+n2
:
C
Let the final temperature be T , then K

E_{1}

+ K

E_{m i x t u r e}

\frac{3}{2}

n_{1} R T_{1}

\frac{3}{2}

n_{2} R T_{2}

\frac{3}{2}

(

n_{1} + n_{2}

) RT
T =

\frac{n_{1} T_{1} + n_{2} T_{2}}{n_{1} + n_{2}}

Question 14. If

C_{1}, C_{2}, C_{3}

.........represent the speeds of

n_{1}, n_{2}, n_{3}

......... molecules, then the root mean square speed is _____.

√n1C12+n2C22+n3C32+...n1+n2+n3+...
⟮n1C12+n2C22+n3C32+..⟯12n1+n2+n3+.....
⟮n1C12⟯12n1+⟮n2C22⟯12n2+⟮n3C32⟯12n3+.....
⟮⟮n1C1+n2C2+n3C3+...⟯2⟮n1+n2+n3+.....⟯⟯12

Discuss Question

Answer: Option A. -> √n1C12+n2C22+n3C32+...n1+n2+n3+...
:
A
Root mean square speed =

\frac{\sqrt{n_{1} {C_{1}}^{2} + n_{2} {C_{2}}^{2} + n_{3} {C_{3}}^{2} + . . .}}{n_{1} + n_{2} + n_{3} + . . .}

Question 15. At what temperature, the volume of 'V' of a certain mass of a gas at

37^{o} C

will be doubled, keeping the pressure constant?

327o C
347o C
527o C
54o C

Discuss Question

Answer: Option B. -> 347o C
:
B

\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}

V_{1} = V, T_{1} = 37 + 273 = 310 k

V_{2} = 2 V, T_{2} = ?

\frac{V}{310} = \frac{2 V}{T_{2}}

T_{2} = 620 K = 347^{o} C

Question 16. For one mole of a van der Waal's gas when b =o and T = 300 K, the PV vs

\frac{1}{V}

plot is shown below. The value of the Van der Waal's constant "a" (

a t m L^{2} m o l^{- 2}

) is

For One Mole Of A Van Der Waal's Gas When B =o And T = 300 K...

Discuss Question

Answer: Option C. -> 1.5
:
C

⟮ P + \frac{a}{V^{2}} ⟯ ⟮ V ⟯ = R T

PV +

\frac{a}{V}

= RT
y = RT –a(V)
y = RT –a(x)
So, slope = a =

\frac{21.6 - 20.1}{3 - 2} = \frac{1.5}{1} = 1.5

Question 17. The compressibility factor of a gas is less than 1 at STP. Its molar volume Vm will be

Vm > 22.42
Vm < 22.42
Vm = 22.42
None

Discuss Question

Answer: Option B. -> Vm < 22.42
:
B
If Z<1 then molar volume is less than 22.4 L

Question 18. If pressure becomes double at the same absolute temperature on 2 L of

C O_{2}

, then the final volume of

C O_{2}

will be?

2 L
4 L
25 L
1 L

Discuss Question

Answer: Option D. -> 1 L
:
D

\frac{P_{1}}{P_{2}}

\frac{1}{2}

\frac{V_{1}}{V_{2}}

\frac{P_{1}}{P_{2}}

\frac{2}{1}

\frac{2 L}{V_{2}}

\frac{1}{2}

;

V_{2}

= 1L

Question 19. Calculate the volume occupied by 16 gram

O_{2}

at 300 K and 8.31 Mpa if

\frac{P_{c} V}{R T_{c}} = \frac{3}{8} a n d \frac{P_{r} V_{r}}{T_{r}} = 2.21 ⟮ : R = 8.314 \frac{M p a}{K - m o l} ⟯

124.31 lit
124.31 mL
248.62 mL
248.62 L

Discuss Question

Answer: Option B. -> 124.31 mL
:
B

\frac{P V_{m}}{R T} = \frac{3}{8} \times 2.21; V_{m} = ⟮ \frac{3}{8} \times 2.21 ⟯ \times \frac{R T}{P}

V_{m} = \frac{3}{8} \times 2.21 \times \frac{8.314 \times 300}{8.314} = 248.625 m L

V_{O_{2}} = \frac{16}{32} \times 248.625 = 124.31 m L

Question 20. The rate of diffusion of Methane at certain temperature is 2 times that of gas X. The molecular mass of X is___.

Discuss Question

Answer: Option A. -> 64
:
A

\sqrt{\frac{M_{x}}{16}} = \sqrt{\frac{r_{C H_{4}}}{r_{x}}} = \sqrt{\frac{2}{1}}

Squaring both sides, we get

M_{x} = 16 \times 4 = 64

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