Question

U SR, test tube, apparatus, cyclohexane 2. Mass of flask, test tube, apparatus 3. Mass of cyclohexane 4. Freezing point, from cooling curve 203.239 g 7.9079 5.9°C B. Freezing Point of Cyclohexane plus Unknown Solute Unknown Solute Trial 1 Trial 2 Trial 3 1. Mass of cyclohexane 1.9029 8.152 8.40ly 2. Mass of solute, total Bolby 8.409 9.5989 3. Freezing point, (cooling curve) 1.8 C 2C 7°C 20° 7.9079 Calculations 1. K, for cyclohexane (pure solvent) 2. Freezing point change, AT: 3. Mass of cyclohexane in solution 4. Moles of solute 5. Mass of solute in solution, total 6. Molar mass of solute 20.0°C/m 9.2 1.9024 101 8.7 1.929! 1.9024 8.164 8 .9094 1.5989! 7. Average molar mass of solute 8. Standard Deviation 7.99 1 kg Trong Show calculations for 'Trial 1 below: ATE=kem m = molalitry mal salute kg solvent AT.

Answer 1

ANSWER:

The freezing point change of any solvent is defined as:

$\Delta T=T_{pure\, solvent}-T_{solution}= K_{f}\times m$

According to data, the freezing point of pure cyclohexane is 5.9 ºC (from cooling curve). Then, for each trial, the freezing point change is

Trial	Freezing point change, ΔTf
1	5.9 ºC - 1.8 ºC = 4.1 ºC
2	5.9 ºC - 0.2 ºC = 5.7 ºC
3	5.9 ºC - 0.7 ºC = 5.2 ºC

then, the molality of the solution could be defined as:

$m=\frac{ \Delta T}{K_{f}}=\frac{moles\, solute}{Kg\, solvent}$

and the moles of solute are:

$moles\, solute=\frac{ \Delta T\times Kg\, solvent}{K_{f}}$

Now we can calculate the moles of solute for each trial:

Trial	ΔTf	Kg solvent	Kf	moles of solute
1	4.1 ºC	7.902x10-3 Kg	20.0 ºC/m	$\frac{ 4.1\, ^{\circ}C\times 7.902x10^{-3}\, Kg}{20.0\, \frac{^{\circ}C.Kg}{mol}}=\mathbf{1.62x10^{-3}\, mol}$
2	5.7 ºC	8.152x10-3 Kg	20.0 ºC/m	$\frac{ 5.7\, ^{\circ}C\times 8.152x10^{-3}\, Kg}{20.0\, \frac{^{\circ}C.Kg}{mol}}=\mathbf{2.32x10^{-3}\, mol}$
3	5.2 ºC	8.401x10-3 Kg	20.0 ºC/m	$\frac{ 5.2\, ^{\circ}C\times 8.401x10^{-3}\, Kg}{20.0\, \frac{^{\circ}C.Kg}{mol}}=\mathbf{2.18x10^{-3}\, mol}$

We can determine the molar mass of solute using the moles of solute and the mass:

$Molar\, mass=\frac{mass\, solute}{moles\, solute}$

For each trial:

Trial	mass solute	moles solute	moles of solute
1	8.16 g	1.62x10-3 moles	$\frac{ 8.16\, g}{1.62x10^{-3}\, mol}=\mathbf{5037.04\, g/mol}$
2	8.409 g	2.32x10-3 moles	$\frac{ 8.409\, g}{2.32x10^{-3}\, mol}=\mathbf{3624.57\, g/mol}$
3	8.598 g	2.18x10-3 moles	$\frac{ 8.598\, g}{2.18x10^{-3}\, mol}=\mathbf{3944.04\, g/mol}$

The average molar mass (MW_av) of solute is

$MW_{av}=\frac{MW\, 1+MW\, 2+MW\, 3}{3}$

$MW_{av}=\frac{5037.04+3624.57+3944.04}{3}=\mathbf{4201.88\, g/mol}$

The standard deviation is

$SD=\sqrt{\frac{\sum \left | MW- MW_{av}\right |^{2}}{n-1}}$

$SD=\sqrt{\frac{(5037.04-4201.88)^{2}+(3624.57-4201.88)^{2}+(3944.04-4201.88)^{2}}{3-1}}$

$SD=\mathbf{\pm 740.70\, g/mol}$