Question

To be able to see a measurable effect, your boiling point elevation must be at least 1.00°C. Knowing that the Kb­ of water is 0.512\frac{^\circ\text{C}\cdot\text{kg}}{\text{mol}}0.512 ∘ C ⋅ kg mol, determine what mass of calcium chloride (in g) is needed to see at least a 1.00^\circ\text{C}1.00 ∘ C boiling point increase in 26.63 mL of water.

Answer 1

Answer : The mass of calcium chloride (in g) needed is, 1.92 grams.

Explanation : Given,

Boiling point of elevation constant
`(K_b)` for water =
`0.512^oC/m`

Mass of water (solvent) =
`Density* Volume=1.00g/mL* 26.63mL=26.63g=0.02663kg`

Molar mass of
`CaCl_2` = 110.98 g/mole

Formula used :

`\Delta T_b=i* K_b* m\\\\\Delta T_b=i* K_b*\frac{\text{Mass of }CaCl_2}{\text{Molar mass of }CaCl_2* \text{Mass of water in kg}}`

where,

`\Delta T_b` = change in boiling point =
`1.00^oC`

i = Van't Hoff factor = 3 (for electrolyte)

`K_b` = boiling point constant for water

m = molality

Now put all the given values in this formula, we get

`1.00^oC=3* (0.512^oC/m)* \frac{\text{Mass of }CaCl_2}{110.98g/mol* 0.02663kg}`

`\text{Mass of }CaCl_2=1.92g`

Therefore, the mass of calcium chloride (in g) needed is, 1.92 grams.