Question

Please Explain Fully

3A. 0.0500 mol of a nonvolatile solute is dissolved in exactly 300 mL of water at 27°C. The osmotic pr essure in mm Hg is (a) 3120 (b) 4.1 (c) 3.1 (d) 0.0054

3B. What is the boiling point in °C of a 0.30 m sugar in water solution? (Kb = 0.52°C/m for water) (a) 0.16 (b) 99.84 (c) 100.16 (d) 101.73

Answer 1

PV = nRT

V   = 300ml = 0.3L

T   = 27+273 = 300K

n   = 0.05 moles

P   = nRT/V

    = 0.05*0.0821*300/0.3   = 4.1atm >>>>>answer

B.

Tb = i*Kb*m

   i = 1 for non volatile liquid

Kb = 0.52⁰C/m

m = 0.3m

Tb = i*Kb*m

             = 1*0.52*0.3   = 0.156⁰C

boiling point = 100+0.16 = 100.16⁰C >>>>>>answer