Question

Determine the mass, in grams, of sucrose (C₁₂H₂₂O₁₁,M = 342.3 g/mol) needed to prepare a solution with a vapor pressure of 41.1 mm Hg. The glucose is dissolved in 738 g of water at 40 °C. The vapor pressure of pure water at 35 °C is 42.2 mmHg.

Answer 1

Relative lowering of vapor pressure = (p° - p_s)/p° = x_B (mole fraction of solute )

Mole of water = 738g/18g/mol = 41mol

(42.2 - 41.1 )/42.2 = n_sucrose/(n_sucrose + n_water) = 0.026066

(1-0.026066)*n_sucrose = 41×0.26066

Moles of sucrose = 1.0973 mol

Mass of sucrose is needed to prepare that solution = number of mol × molar mass

= 1.0973 mol × 342.3g/mol = 375.6087 g

= 375.6 g. (Answer)