1.

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

Answer»

0.001 M
0.1 M
0.02 M
0.01 M

Solution :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`


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