Correct Answer - `a. 0.024M, b. 0.03m`
`a.` `Mw of Co(NO_(3))_(2).6H_(O)=58.7+2(14+48)+6xx18`
`=310.78g mol^(-1)`
`M=(W_(2)xx1000)/(Mw_(2)xxV_(sol)(i n mL))`
`=(30gxx1000)/(310.7gmol^(-1)xx4300mL)`
`=0.02M`
`b`. `M_(1)V_(1)=M_(2)V_(2)`
`0.5xx30mL=M_(2)xx500`
`:M_(2)=0.03M`