Question

3a. How many grams of sucrose must be added to 430g of water to lower the vapor pressure 1.4mmHg at 25C? The vapor pressure of water at 25C is 233.8 mmHg and molar mass is 342.3 g/mol

3. Vapor pressure lowering a) How many grams of sucrose must be added to 430 R of water to lower the vapor pressure 1.4 mm Hg at 25 °C? The vapor pressure of water at 25°C is 23.8 mm Hg and the molar mass of sucrose is 342.3 g/mol (5.1 x 10 g) that I molto l = 23.87altao 1911 how osm ni = 430g 420 na = liemelty b) How many grams of NaCl must be added to the same amount of water as in part a) to lower the vapor pressure the same amount as in a) at 25.0 °C? Assume complete dissociation of NaCl (58.44 g/mol). [42.3 g] 589 T8 X=423 4. An aqueous solution is 20.% glucose by mass (d = 1.078 g/mL at 20°C). Calculate the following: (Some useful constants for water: Kr= 1.86 °C/m and Ko = 0.512 °C/m) a) freezing point [-2.6°C] b) boiling point at 1 atm [100.71 °C] c) osmotic pressure [29 atm] omade mesure

Answer 1

3a)

No. of moles of sucrose = n₁

No. of moles of water = n₂ = (430 g)/(18.02 g/mole) = 23.86 mole

Vapor pressure lowering = Δp= 1.4 mmHg

Vapor pressure of pure water = p⁰ = 23.8 mmHg

Calculate the no. of moles of sucrose using following formula

(Δp/p⁰) = (n₁)/( n₁+ n2)

(1.4/23.8) = (n₁)/( n₁+ 23.86 mole)

(0.0588)(n₁ + 23.86 mole) = n₁

0.0588 n₁ + 1.403 = n₁

n₁ - 0.0588 n₁ = 1.403

0.941n₁ = 1.403

n₁ = 1.403/0.941 = 1.491 mole

No. of moles of sucrose = 1.491 mole

Grams of sucrose = (No. of moles of sucrose)(Molar mass of sucrose)

= (1.491 mole)(342.3 g/mole)

= 510.36 g

Grams of sucrose = 510.4 g