Calculate the molarity of each of the following solutions `:`
`a. 30g` of `Co(NO_(3))_(2).6H_(2)O` in `4.3L` of solution
`b. 30mL` of `0.5 M H_(2)SO_(4)` diluted to `500mL`.

Explanation

(a) Molarity of solution `= ("Mass of solute/Molar mass of solute")/("Volume of solution in litres")`
Mass of solutem `Co(NO_(3))_(2).6H_(2)O=30g`
Molar mass of solute, `Co(NO_(3))_(2).6H_(2)O=59+2xx14+6xx16+6xx18=291 "g mol"^(-1)`
Volume of solution `=4.3 L`
Molarity `(M)=((30g)//(291 "g mol"^(-1)))/((4.3L))=0.024"mol L"^(-1)=0.024M`
(b) Volume of undiluted `H_(2)SO_(4)` solution `(V_(1))=30` mL
Molarity of undiluted `H_(2)SO_(4)` solution `(M_(1))=0.5` M
Volume of diluted `H_(2)SO_(4)` solution `(V_(2))=500` mL
Molarity of diluted `H_(2)SO_(4) (M_(2))` can be calculated as :
`M_(1)V_(1)=M_(2)V_(2)`
or `M_(2)=(M_(1)V_(1))/(V_(2))=((30mL)xx(0.5M))/((500mL))=0.03M`.