Question

When 2.14 g of a nonelectrolyte solute is dissolved in water to make 165 mL of solution at 22 °C, the solution exerts an osmotic pressure of 821 torr.

Answer 1

Solution :-

mass of solute = 2.14 g

volume = 165 ml = 0.165 L

Temperatrue = 22 C +273 = 295 K

osmotic pressure = 821 torr * 1atm / 760 torr = 1.08 atm

lets first calculate the molarity of the solution

= MRT

where = osmotic pressure

M= molarity , R= gas constant , T =kelvin temperature

lets put the values in the formula

M= /RT

M= 1.08 atm / (0.08206 L atm per K.mol * 295 K)

M= 0.0446 M

therefore the molarity of the solution = 0.0446 M

now lets calculate the moles of the solute

moles = molarity * volume in liter

         = 0.0446 mol per L * 0.165 L

        = 0.007359 mol

now lets calculate the molar mass of the solute

molar mass = mass / moles

                 = 2.14 g / 0.007359 mol

                 = 291 g/ mol

Therefore molar mass of the non electrolyte solute = 291 g/mol