Question

When 4.57 g of a nonelectrolyte solute is dissolved in water to make 835 mL of solution at 23 °C, the solution exerts an osmotic pressure of 869 torr. What is the molar concentration of the solution? concentration: How many moles of solute are in the solution? moles of solute: mol What is the molar mass of the solute? molar mass: g/mol

Answer 1

Given

Mass of Solute = 4.57 g

Volume of solution =835 ml = 0.835 L

Osmotic pressure = 869 torr =869 torr x ( 1 atm / 760 torr ) = 1.143 atm

Temperature = 23 + 273 = 296 K

Solute is non electrolyte , therefore i = 1 .

We have relation between osmotic pressure and Molarity of solution as, = i C R T

Where, is osmotic pressure of a solution, i is a van't Hoff factor, C is a concentration in mol / L,T is a temperature of Solution.

Substituting given values in above relation, we get

1.143 atm = C $\times$ 0.082057 L atm mol ^-1 K ^-1$\times$ 296 K

C = 1.143 atm / 0.082057 L atm mol ^-1 K ^-1$\times$ 296 K

Concentration of solution = 0.0471 M

We have, Molarity = No. of moles / Volume of solution in L

0.0471 mol / L = No. of moles of solute / 0.835 L

No. of moles of solute = 0.0471 mol / L $\times$ 0.835 L

= 0.03932 mol

We have, no. of moles = mass / Molar mass

Molar Mass of Solute = Mass / no. of moles of solute

Molar Mass of Solute =4.57 g / 0.03932 mol =116.2 g / mol

ANSWER : Molarity of solution = 0.0471 M

Molar Mass of Solute = 116  g / mol

Moles of solute = 0.0393 mol