(a) Differentiate between molarity and molality for a solution. How does a change in temperature influence their values ?
(b) Calculate the freezing point of an aqueous solution containing `10.50g` of `MgBr_(2)` in `200g` of water. (Molar mass of `MgBr_(2)=184g`) (`K_(f)` for waer `=1.86K kg "mol"^(-1)`)

Answer

(a) Molality: It is the number of the solute dissolve per kilogram `(kg)` of the solvent. It is denoted by
`therefore` Molality `(m)` =("Moles of solute")/("Mass of solvent in"kg)`
`tehrefore m=(w_(2)xx1000)/(M_(2)xxw_(1))`
where `w_(1)=` mass of solvent, `w_(2)=` mass of solute and `M_(2)=` Molar mass of solute.
Molarity: It is the number of moles of the solute dissoved per litre of the solutions. It is denoted by `M`.
`therefore` Molarity `(M)`=("Moles of solute")/("Volume of solution in litre")`
`therefore M=(w_(2)xx1000)/(M_(2)xxV)`
where `w_(2)=` mass of solute, `M_(2)=` molar mass of solute and `V=` Volume of solution.
The molarity and molality of a solution will be nearly same if the mass of solvent is nerarly equal to the volume of solution.
Molarity decreases because volume of the solution increases with increases in temperature. But molaity remains unaffected.
(b) Molecular mass of `MgBr_(2)=184g`
mass of `MgBr_(2)`=184g)`
`DeltaT_(f)=(K_(f)xxW_(B)xx1000)/(M_(B)xxW_(A))=(1.86xx10.50xx1000)/(184xx200)=0.530`
Freezing point of solution `=0-530=-0.530^(@)C.