1.

Weight of Urea required 200ml of 2 M solution will be

Answer»

12 gm
24 gm
20 gm
60 gm

Solution :Molarity `= (W_(B))/(M_(B))xx(1000)/(V_((ml)))`
V = 200 ml
Molarity = 2 M and `M_(B) = 60 g`
`W_(B) = (M_(B)xx V_((ml))xx" Molarity")/(1000)=(60xx200xx2)/(1000)=24 gm`.


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