A mixture of solid NaCl and CaCl₂ contains 10.0 g of NaCl, and an unknown amount of CaCl₂. When the entire mixture is dissolved in 1.00 kg of water, the boiling point changes by 0.25 °C. What mass of CaCl₂ was in the original mixture?
K_b for water is 0.512 K^. kg ^. mol^-1.

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Quizplus · Accepted Answer

The change in boiling point is given by:ΔT_b = K_b * m * iwhere K_b is the boiling point elevation constant for water, m is the molality of the solution, and i is the van't Hoff factor.The van't Hoff factor for NaCl is 2, and the van't Hoff factor for CaCl₂ is 3.The molality of the solution is given by:m = (moles of solute) / (kg of solvent)The moles of NaCl in the solution are:moles of NaCl = 10.0 g / 58.44 g/mol = 0.171 molThe moles of CaCl₂ in the solution are:moles of CaCl₂ = (ΔT_b / K_b * m * i) - moles of NaClmoles of CaCl₂ = (0.25 °C / 0.512 K^. kg ^. mol^-1 * 1.00 kg * 3) - 0.171 molmoles of CaCl₂ = 0.116 molThe mass of CaCl₂ in the original mixture is:mass of CaCl₂ = 0.116 mol * 110.98 g/mol = 12.9 gTherefore, the mass of CaCl₂ in the original mixture is 12.9 g.

A Mixture of Solid NaCl and CaCl2 Contains 10

A Mixture of Solid NaCl and CaCl₂ Contains 10