Question

When the cells of the skeletal vacuole of a frog were placed in a series of NaCl solutions of different concentrations at 298 K, it was found that the cells remained unchanged in a 0.70 % (by weight) NaCl solution. For a 0.70 % NaCl solution the freezing point is depressed by –0.406 K. The osmotic pressure was calculated to be 5.34 atm.

a. Suppose that sucrose (M=342 g/mol) was used instead of NaCl (M= 58.5 g/mol) to make the isoosmotic solution. Estimate the concentration (wt%) of sucrose that would be sufficient to balance the osmotic pressure of the cytoplasm of the cell. Assume that sucrose solutions behave ideally.

Answer 1

Answer – We are given, osmotic pressure of the cytoplasm of the cell = 5.34 atm,

Molar mass of sucrose = 342 g/mol

isosmotic solution, so the freezing point is depressed = -0.406 K

we know, the formula for osmotic pressure

π = MRT

where , M = moles / L

so, π = nRT/V

where , molality of solute = n solute / weight of solvent in kg

freezing point is depressed

ΔTf = -Kf * m

So, m = - ΔTf / Kf

We also know, density = mass / volume

So, π = (- ΔTf / Kf) RT / (1/density )

We know density of water = 1.0 g/mL

So in the kg/L = 1000 kg/L

5.34 atm = (- ΔTf / Kf) * 0.0821 L.atm.mol^-1.K^-1 * 298 K / (1/1000 kg/L)

5.34 * 0.001 L/kg = (- ΔTf / Kf) * 24.46

So, (- ΔTf / Kf) = 0.00534 / 24.46

                          = 0.000218

So, molality = (- ΔTf / Kf) = 0.000218 m

So, molality = moles / kg of solvent

So, moles = 0.000218 m * 1.0 kg

                = 0.000218 moles

So mass of sucrose = 0.000218 moles * 342 g/mol

                               = 0.0746 g

Weight percent of sucrose = 0.0746 g / 1000 g * 100 %

                                            = 0.0074 %