Question

How many grams of sucrose (C_12 H_22 O_11, 342.3 g/mol) would be needed to make 2.5 L of a solution with an osmotic pressure of 14 atm at 25 degree C? (R = 0.0821 L middot atm/K middot mol) A) 0.57 g B) 6.8 g C) 2.0 times 10^2 g D) 4.9 times 10^2 g E) 5.8 times 10^3 g

Answer 1

The answer is D) 4.9×10^2g

Osmotic pressure , $\prod$ = MRT

Where, M = molarity of solute

R = gas constant ,0.0821(Lit atm / mol K)

T = Temperature in K , 298K

Therefore,

14 atm = M × 0.0821(Lit atm / mol K ) × 298K

M = 14 atm /( 0.0821(Lit atm / mol K) × 298K )

= 0.5722M

Therefore, Molarity of sucrose = 0.5722M = 0.5722 mol /Lit

For 2.5Lit sucrose solution no of mole of sucrose need= 0.5722(mol/Lit)× 2.5Lit = 1.4305mol

Molar mass of Sucrose = 342.3g/mol

Mass of Sucrose = No of mole × molar mass

= 1.4305 × 342.3g = 489.66g = 4.9×10^2g