1.

In a flask, the weight ratio of CH_(4) (e) and SO_(2) (g) at 298 K and I bar is 1 : 2. The ratio of the number of molecules of SO_(2) (g) and CH_(4) (g) is

Answer»

`1:4`
`4:1`
`1:2`
`2:1`

SOLUTION :Let mass of `CH_(4)(G) = 1g`
Number of moles of `CH_(4)(nCH_(4)) =1/16`
Number of MOLECULES of `CH_(4)(g) = 1/16 xx N_(A)`
Let the mass of `SO_(2)(g) = 2g`
Number of moles of `SO_(2)(g) (n_(SO_(2))) = 2/64`
Number of molecules of `SO_(2)(g) = 2/64 xx N_(A)`
Ratio of number of molecules of `SO_(2)` (e) and number of molecules of `CH_(4)`(g)
`=2/64 xx N_(A) : 1/16 xx N_(A) rArr 1/32 : 1/16 rArr 1:2`


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