Question

What is the osmotic pressure of a solution made by dissolving 75.0 g of glucose, C6H12O6, in enough water to form 250.0 mL of solution at 25.0 ∘C ?

Answer 1

Molar mass of C6H12O6 = 6*MM(C) + 12*MM(H) + 6*MM(O)

= 6*12.01 + 12*1.008 + 6*16.0

= 180.156 g/mol

mass of C6H12O6 = 75.0 g

we have below equation to be used:

number of mol of C6H12O6,

n = mass of C6H12O6/molar mass of C6H12O6

=(75.0 g)/(180.156 g/mol)

= 0.4163 mol

volume , V = 250 mL

= 0.25 L

we have below equation to be used:

Molarity,

M = number of mol / volume in L

= 0.4163/0.25

= 1.665 M

T= 25.0 oC

= (25.0+273) K

= 298 K

we have below equation to be used:

P = C*R*T

P = 1.665*0.0821*298.0

P =40.7 atm

Answer: 40.7 atm