What is the boiling point of a solution made by mixing 100.0g of CaCO3 with 750.0g of water?
To solve for the boiling point of the solution, we should use the formula for the boiling point elevation:
`Delta T_b = i * K_b * m`
`Delta T_b` = the boiling point elevation [T(solution) - T (solvent)]
`K_b` = ebullioscopic constant (for water = 0.512)
`m` = molality of the solution (moles of solute per kilogram of solvent)
= `((100 g CaCO_3) *(1 mol e CaCO_3)/(100.0869g CaCO_3))/(0.750kg)`
= 1.332 m
`i` = van't Hoff factor (for CaCO3, i = 1)
We can substitute the values in the equation above:
`T_(solution) - T_(solvent) = (1)(0.512)(1.332) = 0.682`
- The boiling point of pure water is 100 degrees Celsius
`T_(solution) = 100 + 0.682 = 100.682 ^o C`
Calcium Carbonate is slightly soluble in water. If it is treated as an ionic compound, the van't Hoff factor will change.
`i` = 2; CaCO3 will dissociate into `Ca^(2+) ` and `CO_3 ^(2-) ` ions.
`T_(solution) - T_(solvent) = (2)(0.512)(1.332) = 1.364 `
`T_(solution) = 100 + 1.364 = 101.364 ^o C `