`6.0xx10^(20)` molecules of urea are present in 100 L of his solution. The concentration of urea solution is

Options

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

Answer

D

Explanation

No. of moles of urea `= (6.02xx10^(20))/(6.02xx10^(23))=10^(-3)` mol
`:. 10^(-3)` mol of urea are present in 100 mL
`:.` Moles of urea present in 1L (1000 mL)
`=10^(-3)xx10=10^(-2)`
`:.` Molarity of solution `= 10^(-2) M = 0.01 M`