Given- Vapour pressure of water, = 17.535 mm Hg

= 25 g of glucose

= 450g of water

Molar mass of water, H₂O = 1 + 1 + 16 = 18 g mol^-1

Molar mass of glucose, C₆H₁₂O₆ = (12×6) + (1×12) + (16×6) = 180 g mol^-1

Using Raoult’s law for solution of non-volatile solute,

→ Equation 1

where is the mole fraction of the solute

Substituting the value of in equation 1, we get,

→

Thus, the vapour pressure of water at 293 K at the given conditions is 17.437 mm Hg