Question

What is the boiling point of a solution prepared by adding 435.0 g of calcium perchlorate (Ca(Cl4)2, 283.98 g/mol) to 285 mL of water? The boiling point constant for water is 0.520 degrees Celcius/m.

Answer 1

Mass of Ca(ClO₄)₂ = 435.0g

Molar mass of Ca(ClO₄)₂ = 238.98 g/mol

Thus, moles of  Ca(ClO₄)₂ = mass/molar mass = 435 g/ (238.98 g/mol) = 1.82 mol

Volume of water = 285mL

Let density of water = 1g/ml

Thus, mass of water = Volume* density = 285 mL * 1g/mL = 285g = 0.285kg

Molality = moles of Ca(ClO₄)₂/ volume of water (in kg) = 1.82mol/ 0.285kg = 6.39m

Boiling of water (T_w) = 100^oC

Let boiling point of solution be T_s

K_b =0.520^oC/m

Now, from elevation in boiling point (T )

T = iK_bm

Where i = van't hoff factor = number of ions produced = 3

Thus, T_s - T_w = 3* 0.520^oC/m * 6.39m = 9.96 ^oC

T_s - 100^oC =  9.96 ^oC

Ts = 109.96^oC

THus, boiling point of a solution is 109.96 ^oC