Calculate the vapor pressure at 50°c of a coolant solution that is 52.0: 48.0 ethylene glycol-to-water by volume. at 50.0°c, the density of water is 0.9880 g/ml, and its vapor pressure is 92 torr. the vapor pressure of ethylene glycol is less than 1 torr at 50.0°c.

If we use 100 mL of solution:
V(ethylene glycol - C₂H₆O₂) = 0,52 · 100 mL = 52 mL.
V(water) = 0,48 · 100 mL = 48 mL.
m(C₂H₆O₂) = 52 mL · 1,115 g/mL = 57,98 g.
n(C₂H₆O₂) = 57,98 g ÷ 62,07 g/mol = 0,934 mol.
m(H₂O) = 48 mL · 0,988 g/mL = 47,424 g.
n(H₂O)  = 45,45 g ÷ 18 g/mol = 2,635 mol.
mole fraction of solvent: 2,635 mol / (2,635 mol + 0,934 mol) =0,73.
Raoult's Law: p(solution) = mole fraction of solvent · p(solvent).
p(solution) = 0,73 · 92 torr = 67,33 torr.

Sorry i dont understand the question can you rephase it?