Question

A solution is prepared by dissolving .074 g of glucose, a molecular solid with the formula C6H12O6 in 927 mL of water at 20 celcius - use Raoult's law to determine the water vapor pressuring lowering at 20 celcius where the vapor pressre and density of pure water are 17.5424 torr and 998.2 g/cm^3, respectively.

-calculate the freezing point decrease for this solution compared to pure water

-calculate the boiling point increase for this solution compared to pure water

-calculate the osmotic pressure of this solution assuming no volume change when the solution is prepared

-considering the results from the above, which colligative property would you use to determine the formula weight of glucose (assuming it was unknown) if your dicision was based only on the relative magnitude of the physical property change.

Answer 1

We have 0.74 g glucose (C₆H₁₂O₆) in 927 mL water at 20⁰C. The density of water is 998.2 kg/m³ (you got the units wrong).

Find out the volume of water in m³ first. We have 927 mL water and 1 m³ = 1.0*10⁶ mL; therefore, mass of 927 mL water = (volume of water)*(density of water) = (927 mL)*(1 m³/1.0*10⁶ mL)*(998.2 kg/m³) = 0.92533 kg = (0.92533 kg)*(1000 g/1 kg) = 925.33 g.

Molar mass of water, H₂O = (2*1.008 + 1*15.9994) g/mol = 18.0154 g/mol.

Molar mass of glucose, C₆H₁₂O₆ = (6*12.01 + 12*1.008 + 6*15.9994) g/mol = 180.1524 g/mol.

Moles of glucose corresponding to 0.74 g glucose = (0.74 g)/(180.1524 g/mol) = 4.1076*10^-3 mole.

Moles of water corresponding to 925.33 g water = (925.33 g)/(18.0154 g/mol) = 51.3633 mole.

Find the mole fraction of the solvent = (moles of water)/(moles of water + moles of glucose) = (51.3633 mole)/(51.3633 + 4.1076*10^-3) mole = 0.9999

Use Raoult’s law to find the vapor pressure of the solution as

P_soln = (mole fraction of solvent)*P⁰_solvent where P⁰_solvent is the vapor pressure of the pure solvent at the said temperature.

Plug in values

P_soln = (0.9999)*(17.5424 torr) = 17.5406 torr (ans).

The freezing point of water is 0⁰C = 273 K.

Find the molality of the solution as molality = (moles of solute)/(kg of solvent) = (4.1076*10^-3 mole)/(0.92533 kg) = 4.43906*10^-3m.

Use the relation ΔT_f = k_f*m where k_f = 1.86⁰C/m is the freezing point depression constant (constant for a particular solvent) of water and ΔT_f is the depression (lowering) in freezing point of water. Plug in values.

ΔT_f = (1.86⁰C/m)*(4.43906*10^-3m) = 8.2566*10^-3 ⁰C.

Now, ΔT_f = (T_f)_pure – (T_f)_soln

==> (T_f)_soln = (T_f)_pure – ΔT_f = (0⁰C) – (0.0082566⁰C) = -0.0082566 ⁰C ≈ -0.00826⁰C (ans)

The boiling point of pure water is 100⁰C

We have already determined the molality of the solution before. Use the relation

ΔT_b = k_b*m where ΔT_b is the elevation in boiling point of the solvent and k_b = boiling point elevation constant (again constant for a particular solvent) of water = 0.51⁰C/m.

ΔT_b = (0.51⁰C/m)*(4.43906*10^-3m) = 0.0022639⁰C

Now, ΔT_b = (T_b)_soln – (T_b)_pure

==> (T_b)_soln = (T_b)_pure + ΔT_b = (100⁰C) + (0.0022639⁰C) = 100.0022639 ⁰C ≈ 100.00223⁰C (ans)

The osmotic pressure of the solution is assumed to be the vapor pressure of the solution. Convert the value to atm first.

The osmotic pressure of the solution is 17.5406 torr = (17.5406 torr)*(1 atm/760 torr) = 0.02308 atm.

Use the relation π = M*R*T where π = osmotic pressure of the solution, R = gas constant and T = absolute temperature; M = molarity of the solution.

Plug in values

0.02308 atm = M*(0.082 L-atm/mol.K)*(273 + 20) K = M*24.026 L-atm/mol

====> M = (0.02308 atm)/(24.026 L-atm/mol) = 0.0009606 mol/L ≈ 9.61*10^-4 M (1 mol/L = 1 M).

Therefore, the molar concentration of the solution is 9.61*10^-4 M (ans).