When 4.71 g of a nonelectrolyte solute is dissolved in water to make 985 mL of solution at 29 °C, the solution exerts an osmotic pressure of 941 torr.
1)What is the molar concentration of the solution?
2)How many moles of solute are in the solution?
3)What is the molar mass of the solute?
use the direct formula
Pi = i x M x R x T
where i= vant half facter = 1 for non electrolyte solution
M = Molarity of the solution
Pi = osmatic pressure = 941 torr convert in to atm = 1.24 atm
R = gas constant = 0.0821 L atm/mol-K
T = 29 ºC = 273 +29 = 302 K
1.24 atm = 1 x M x 0.0821 x302 K
M = 1.24 / 24.79
M = 0.05M
Part-B
no ofmoles = Molarity x volume in liters
M = 0.05 , V = 985 mL = 0.985L
= 0.05 mol/L x 0.985 L
= 0.0492 moles
Part C
use the formula
moles = weight / molar mass
molar mass = weight /moles
= 4.71 g / 0.0492 moles
= 95.61 g/mole
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