An aqueous solution containing 100.0 g of nacl and 100.0 g of cacl2 has a volume of 1.00 l and a density of 1.15 g/ml. the vapor pressure of pure water at 25 ∘c is 23.8 mmhg, and you can assume complete dissociation for both solutes

Missing question: What is the vapor pressure of the solution at 25°C?
n(NaCl) = 100 g ÷ 58,4 g/mol.
n(NaCl) = 1,71 mol. 
NaCl → Na⁺ + Cl⁻, amount of ions are 2 · 1,71 mol = 3,42 mol.
n(CaCl₂) = 100 g ÷ 111 g/mol = 0,9 mol.
CaCl₂ → Ca²⁺ + 2Cl⁻, amount of ions 3 · 0,9 mol = 2,7 mol.
m(solution) = 1000 ml (1,00 L) · 1,15 g/ml = 1150 g.
m(H₂O) = 1150 g - 100 g - 100 g = 950 g.
n(H₂O) = 950 g ÷ 18 g/mol = 118,75 mol.
water's mole fraction = 118,75 mol ÷ (118,75 mol + 2,7 mol + 3,42 mol).
water's mole fraction = 0,95.
p(solution) = 0,95 · 23 mmHg = 21,85 mmHg.

The answer is 65.3312