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a gas has vapour density 11.2 the volume occupied by 1g of gas at NTP is a) 1L b) B)11.2L c)22.4L D)4L

princess payal , 12 Years ago
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Yogita Nain

Last Activity: 12 Years ago

molecular mass of gas = 2*V.D.

                                 =2*11.2 =22.4

no of moles of gas = weight/ molecular mass

                            = 1/22.4

vol oF gas = no of moles * V

                = 1/ 22.4 * 22.4

                = 1 L

Godfrey Classic Prince

Last Activity: 12 Years ago

Dear princess payal,

The answer is a) 1L ..

 

All the Very Best & Good Luck to you..

Hope this helped you immensely..

 

Regards,

AskIITians Expert,

Godfrey Classic Prince

IIT-M

 

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aravind

Last Activity: 9 Years ago

its A 
SINCE VAPOUR DENSITY OF A GAS  * 2 =MOLECULAR WEIGHT OF THA GAS 
M.WT.=2*V.D 
          = 2*11.2
          = 22.4g
therefore 22.4g 0f gas at ntp occupies 22.4 L volume 
then 1g oggas at ntp occupies 1L 
 

ghg

Last Activity: 7 Years ago

The answer to your question is 1 litre.
This is because VD=2*Molecular mass
 
SINCE VAPOUR DENSITY OF A GAS  * 2 =MOLECULAR WEIGHT OF THA GAS 
M.WT.=2*V.D 
          = 2*11.2
          = 22.4g
therefore 22.4g 0f gas at ntp occupies 22.4 L volume 
then 1g oggas at ntp occupies 1L 
 

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