`6.02xx10^(20)` molecules of urea are present in 100 ml of its solution. The concentration of solution is :

Options

A. `0.02M`
B. `0.01M`
C. `0.001M`
D. `0.1M`

Answer

B

Explanation

`"Number of moles"=(6.02xx10^(20))/(N_(2))`
`=(6.02xx10^(20))/(6.02xx10^(23))=1xx10^(-3)mol`
Volume of the solution = 100 mL = 0.1 L
Concentration of urea soultion `("in mol "L^(-1))`
`=(1xx10^(-3))/(0.1)=0.01"mol "L^(-1)`