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) = $Density\times Volume=1.00g/mL\times 26.63mL=26.63g=0.02663kg$

Molar mass of $CaCl_2$ = 110.98 g/mole

Formula used :  

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

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

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

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

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