1.

120 gm of urea are present in 5 litre solution, the active mass of urea is

Answer»

`0.2`
`0.06`
`0.4`
`0.08`

SOLUTION :Active `"MASS"=("MOLES")/("Litre")`
`=("wt. in gm/molecular wt.")/("V in litre")=(120//60)/(5)=2/5=4`


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