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The volume of neon gas in "cm"^(3) at ST...

The volume of neon gas in `"cm"^(3)` at STP having the same number of atoms as that present in 800 mg of Ca is (At. Mass : Ca = 40 g mol`""^(-1)`, Ne = 20 g mol`""^(-1)` )

A

56

B

896

C

224

D

448

Text Solution

Verified by Experts

Mass of Ca = 800mg `= (800)/(1000) =0.8` g
Moles of Ca in `0.8` g `= (0.8)/(40) = 2xx 10^(2)`
`because` At STP, volume of 1 mole `= 22400 "cm"^(3)`
`therefore` At STP, volume of `2 xx 10^(-2)` moles
`= 22400 xx 2 xx 10^(-2) = 448 "cm"^(3)`
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