Weight of urea requied to prepare 200 mL of 2M solution as :

Options

A. 12 g
B. 24 g
C. 20 g
D. 60 g

Answer

b

Explanation

`((0.2 molL^(-1))xxW)/((60g mol^(-1))xx(0.2L))`
`W=(0.2 mol L^(-1))xx(60g mol^(-1))xx(0.2L)`
=24g.