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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.

Respuesta :

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

Explanation : Given,

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

Mass of water (solvent) = [tex]Density\times Volume=1.00g/mL\times 26.63mL=26.63g=0.02663kg[/tex]

Molar mass of [tex]CaCl_2[/tex] = 110.98 g/mole

Formula used :  

[tex]\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}}[/tex]

where,

[tex]\Delta T_b[/tex] = change in boiling point  = [tex]1.00^oC[/tex]

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

[tex]K_b[/tex] = boiling point constant for water

m = molality

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

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

[tex]\text{Mass of }CaCl_2=1.92g[/tex]

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

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