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Standard XII

Chemistry

Percentage Composition

The weight of...

Question

The weight of $C H_{4}$ in 9-L cylinder at $27^{0} C$ temperature and $16$ atm pressure is ( $R = 0.08$ L atm $K^{- 1} m o l^{- 1}$ ).

9.6

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96.0

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4.8

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48.0

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Solution

The correct option is A $96.0$ g
Using the ideal gas equation,
$P V = n R T$
$16 a t m \times 9 L = n \times 0.082 \times 300 K$ .....(given)
Therefore, n=6
$n = \frac{W e i g h t}{a t o m i c m a s s}$
weight of $C H_{4} = 16 g {m o l}^{- 1} \times 6 m o l = 96 g$

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Similar questions

Q. Total pressure of a gaseous mixture containing

28 g N_{2}

and

16 g O_{2}

in a cylinder of volume

8.21 L

300 K

is:
[Given:

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

]

Q. Assume that decomposition of

H N O_{3}

can be represented by the following equation

4 H N O_{3} (g) ⇌ 4 N O_{2} (g) + 2 H_{2} O (g) + O_{2} (g)

and the reaction approaches equilibrium at

400 K

temperature and

30 a t m

pressure. At equilibrium partial pressure of

H N O_{3}

2 a t m

.
Calculate

K_{c}

{(m o l / L)}^{3}

400 K

:(Use :

R = 0.08 a t m - L / m o l - K

)

Q. Calculate the weight of

O_{2} (g)

necessary to fill up a cylinder of 5

L

capacity at

27 ° C

and 100

atm

pressure, when the compressibility factor is 0.96.
(Take

R = \frac{1}{12} {L atm mol}^{- 1} K^{- 1}

)

Q. At what pressure and

127^{o}

C, the density of

O_{2}

gas becomes

1.6

g/L? [

a = 4.0

atm

L^{2}

m o l^{- 2}

0.4

m o l^{- 1}

, R

= 0.08

L-atm /K-mol]

Q. Assume that the decomposition of

{H N O}_{3}

can be represented by the following equation

4 {H N O}_{3} (g) ⇌ 4 {N O}_{2} (g) + 2 H_{2} O (g) + O_{2} (g)

The reaction approaches equilibrium at

400 K

temperature and

30

atm pressure. At equilibrium partial pressure of

{H N O}_{3}

2

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Calculate

K_{c}

{(m o l / L)}^{3}

400 K

:
(Use :

R = 0.08 a t m - L / m o l - K

)

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The weight of CH4 in 9-L cylinder at 270C temperature and 16 atm pressure is (R=0.08 L atm K−1mol−1).

The correct option is A 96.0 gUsing the ideal gas equation,PV=nRT16atm×9L=n×0.082×300K .....(given)Therefore, n=6n=Weightatomicmassweight of CH4=16gmol−1×6mol=96g

The weight of $C H_{4}$ in 9-L cylinder at $27^{0} C$ temperature and $16$ atm pressure is ( $R = 0.08$ L atm $K^{- 1} m o l^{- 1}$ ).

The correct option is A $96.0$ g
Using the ideal gas equation,
$P V = n R T$
$16 a t m \times 9 L = n \times 0.082 \times 300 K$ .....(given)
Therefore, n=6
$n = \frac{W e i g h t}{a t o m i c m a s s}$
weight of $C H_{4} = 16 g {m o l}^{- 1} \times 6 m o l = 96 g$