Which has maximum volume at $S T P$ ? ( $1$ mole of a gas $= 22.4 L$ at $S T P$ )

1.6 g

C H_{4}

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6.02 \times 10^{22} N_{2}

molecules

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4.4 g

C O_{2}

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Equal

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Solution

The correct option is D Equal
$1.6 g$ $C H_{4} = 0.1$ mole.
$6.02 \times 10^{22} N_{2}$ molecules $= 0.1$ mole.
$4.4 g$ $C O_{2} = 0.1$ mole.
So, they all have same volume $2.24$ L.

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Similar questions

8

mg of

C H_{4}

gas is removed from

6.02 \times 10^{20}

molecules of

C H_{4}

, then the volume occupied by

C H_{4}

gas left at

S T P

is:

One gram mole of a gas at STP occupies 22.4L volume. This fact was derived from –

When 22.4 L of C $_{4}$ H $_{8}$ at STP is burnt completely, 89.6 L of CO $_{2}$ gas at STP and 72 g of water are produced. The volume of the oxygen gas at STP consumed in the reaction is closest to :

Volume of 4.4g of $C O_{2}$ at STP is

Q. Amount of gas which occupies

22.4 L

volume at STP is equal to vapor density of the gas.

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Which has maximum volume at STP ? (1 mole of a gas =22.4L at STP)

The correct option is D Equal1.6g CH4=0.1 mole.6.02×1022N2 molecules =0.1 mole.4.4g CO2=0.1 mole.So, they all have same volume 2.24 L.

Which has maximum volume at $S T P$ ? ( $1$ mole of a gas $= 22.4 L$ at $S T P$ )

The correct option is D Equal
$1.6 g$ $C H_{4} = 0.1$ mole.
$6.02 \times 10^{22} N_{2}$ molecules $= 0.1$ mole.
$4.4 g$ $C O_{2} = 0.1$ mole.
So, they all have same volume $2.24$ L.