#### Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Click to Chat

1800-5470-145

+91 7353221155

CART 0

• 0
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

# 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

Yogita Nain
5 Points
9 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
633 Points
9 years ago

Dear princess payal,

The answer is a) 1L ..

All the Very Best & Good Luck to you..

Hope this helped you immensely..

Regards,

Godfrey Classic Prince

IIT-M

Please approve my answer if you liked it by clicking on "Yes" given below..!!

aravind
23 Points
5 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
11 Points
3 years ago
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