Calculate relative lowering of vapor pressure of 1 molal glucose solution in water at room temperature. (molar mass of water =18 )

Step-by-step solution

A 1 molal solution of glucose means 1 mole of glucose is dissolved in 1000 g of water. Number of moles of solute (glucose), n = 1 mol Number of moles of solvent (water), N = 1000 18 = 55.55 mol According to Raoult's law, the relative lowering of vapor pressure is equal to the mole fraction of the solute. P P^ = _ solute = n n + N P P^ = 1 1 + 55.55 = 1 56.55 P P^ 0.01768 1.8 10⁻² Answer: 1.8 10⁻²