`18g` of glucose (molar mass `180g "mol"^(-1)`) is present in `500CM^(3)` of its aqueous solution. What is the molarity of the solution? What additional data is required if the molality of the solution is also required to be calculated?

Explanation

Given Mass of solute (glucose) `=18g`
Molar mass of the solute (glucose) `=180g "mol"^(-1)`
Volume of solution `=500cm^(3)`
Asked Molarity of solution `(M)=?`
Formulae: `M=(W_(B))/(M_(B))xx(1000)/("Volume of solution in" cm^(3))`
Explanation: `M=` molarity of solution, `W_(B)=` Mass of solute, `M_(B)=` Molecular weight solute
Subsitution `&` Calculation
`M=(18)/(180)xx(1000)/(500cm^(3))=0.2"mol" L^(-1)`.