Question

What amount of sucrose (C12H22O11) should be added to 5.83 mol water to lower the vapor pressure of water at 50 °C to 72.0 torr? The vapor pressure of pure water at 50 °C is 92.6 torr. ​

Answer 1

moles of water = n2 = 5.83

moles of sucrose = n1

relative lowering vapour pressure = mole fraction of solute

Po - Ps / Po = n1 / n1 + n2

92.6 - 72.0 / 92.6 = n1 / n1 + 5.83

0.2225 = n1 / n1 + 5.83

0.2225 n1 + 1.297 = n1

1.297 = 0.7775 n1

n1 = 1.668

moles of sucrose = 1.668

mass of sucrose = 1.668 x 342

                           = 570. g