Calculate the molarity of each of the following solution :
(a) 30g of `CO(NH_(3))_(2).6H_(2)O` in 4.3 L of solution.
(b) 30 mL of `0.5" MH"_(2)"SO"_(4)` diluted to 500 mL.
(a) Molarity `=("moles of solute")/("Volume of solution litre")`
and moles of solute `=("mass of solute")/("molar solution of solute")`
So, first find molar mass by adding atomic masses of different elements, then find moles of solute and then molarity.
(b) Use molarity equation for dilution.
`M_(1)V_(1)=M_(2)V_(2)`
(Before dilution) (After dilution)

Explanation

Molar mass of `CO(NO_(3))_(2).6H_(2)O=(58.7)+2(14+48)+(6xx18)"g mol"^(-1)`
`=58.7+124+108=290.7~~291" g mol"^(-1)`
Mole of `CO(NO_(3))_(2)6" H"_(2)O=(30"g")/(291" g mol"^(-1))=0.103`
Volume of solution = 4.3 L
Molarity (M) `=(0.103" mol")/(4.3" L")=0.024" mol L"^(-1)=0.024" M"`
(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"`
We know that `M_(1)V_(1)=M_(2)V_(2)`
`:. M_(2)=(M_(1)V_(1))/(V_(2))=((0.5"M")(30" mL"))/(500" mL")=0.03" M"`