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

Explanation

`"Molarity of solution"=("Mass of solute/Molar mass of solute")/("Volume of solution in litres")`
Molar mass solute, `Co(NO_(3))_(2).6H_(2)O=30 g`.
Molar mass of solute, `Co(NO_(3))_(2).6H_(2)O=59+2xx14+6xx16+6xx18=219 g mol^(-1)`
Volume of solution=4.3 L
`"Molarity(M)"=((30g)//(291g mol^(-1)))/((4.3L))=0.024 molL^(-1)=0.024 M`
Volume of undiluted `H_(2)SO_(4)` solution `(V_(1))`=30 mL
Molarity of undiluted `H_(2)SO_(4)` solution `(M_(1))`=1.5 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)`
`M_(2)=(M_(1)V_(1))/V_(2)=((30mL)xx(0.5M))/((500mL))=0.03 M`