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Standard XII

Chemistry

Percentage Composition

A gas cylinde...

Question

A gas cylinder was found unattended in a public place. The investigating team took the collected samples from it. The density of the gas was found to be $2.380 g L^{- 1}$ at $15^{o} C$ and pressure. Hence the molar mass of the gas is:

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Solution

The correct option is A 58
Applying ideal gas law,
$P V = n R T$
$P = \frac{w t}{M w t \times V o l u m e} \times R T$

$M w t = \frac{d}{P} \times R T = \frac{2.380}{1} \times 0.0821 \times 288 = 58$

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A gas cylinder was found unattended in a public place. The investigating team took the collected samples from it. The density of the gas was found to be 2.380 gL−1 at 15oC and pressure. Hence the molar mass of the gas is:

The correct option is A 58Applying ideal gas law,PV=nRTP=wtMwt×Volume×RTMwt=dP×RT=2.3801×0.0821×288=58

A gas cylinder was found unattended in a public place. The investigating team took the collected samples from it. The density of the gas was found to be $2.380 g L^{- 1}$ at $15^{o} C$ and pressure. Hence the molar mass of the gas is:

The correct option is A 58
Applying ideal gas law,
$P V = n R T$
$P = \frac{w t}{M w t \times V o l u m e} \times R T$

$M w t = \frac{d}{P} \times R T = \frac{2.380}{1} \times 0.0821 \times 288 = 58$