Question

When 2.53 g of a nonelectrolyte solute is dissolved in water to make 995 mL of solution at 25 °C, the solution exerts an osmotic pressure of 815 torr. What is the molar concentration of the solution? How many moles of solute are in the solution? What is the molar mass of the solute?

Answer 1

a)

P= 815.0 torr

= (815.0/760) atm

= 1.0724 atm

T= 25.0 oC

= (25.0+273) K

= 298 K

we have below equation to be used:

P = C*R*T

1.0724 = C*0.0821*298.0

C =0.0438 M

Answer: 0.0438 M

b)

volume , V = 995 mL= 0.995 L

we have below equation to be used:

number of mol,

n = Molarity * Volume

= 0.0438*0.995

= 4.361*10^-2 mol

Answer: 4.361*10^-2 mol

c)

mass of solute = 2.53 g

we have below equation to be used:

number of mol = mass / molar mass

4.361*10^-2 mol = (2.53 g)/molar mass

molar mass = 58.01 g/mol

Answer: 58.01 g/mol