If the vapor pressure of the pure solvent at room temperature is 4860 torr what

# If the vapor pressure of the pure solvent at room

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solvent. If the vapor pressure of the pure solvent at room temperature is 48.60 torr, what will be the vapor pressure (in torr) of the solution described at the same temperature? χ solvent=moles of water ÷ (moles of solute + moles of solvent) χ solvent = 34.15 ÷ (8.450+34.15)= 0.8016 χ solvent= 0.8016 Therefore, the pressure of the solution is then calculated via Raoult's Law: Psolution=0.8016×48.60=38.96torr
Jenele McLean 3 6. t-Butanol has a freezing point of 25.82 o C and a freezing point depression constant of 8.37 o C/ m . What will be the freezing point (in o C) of a solution that is 0.207 molal solute dissolved in t-butanol? T f = (Freezing point of pure solvent) - (Freezing point of solution) T f = (K f ) (m) T f = (8.37 o C/ m x 0.207m) =1.732 0 C 1.732= 25.82 o C – (X) X=25.82-1.732 24.088 Freezing point of the solution= 24.09 o C 7. What is the osmotic pressure (in atm) of a 0.225 M solution of glucose at 315 K? Osmotic pressure is expressed by the formula: w 2 = Mass of the solute in gram M 2 = Molar mass of the solute in gram/mol V = Volume of the solution in Liter R = the universal gas constant T = Temperature in Kelvin Glucose= C6H12O6 (12x6)+(1x12)+(16x6)=180g/mol 1M solution of glucose contains 180 g of glucose in 1L solution Therefore: o.225M solution of glucose contains 180gx0.225= 40.5g of glucose in 1L solution W 2 = 40.5g M2= 180g/mol V = 1L
Jenele McLean 4 R = 0.0821 L atm/K mol T= 315K (40.5g x 0.0821 L atm/K mol x 315K) ÷ (180X1L) = 5.819 atm 8. A 0.608 g sample of a protein was dissolved in 25.00 mL of water and the resulting osmotic pressure at 305 K was measured at 4.15 x 10 -4 atm. What is the molar mass of this protein? W2= 0.608g M2=?? V= 25 mL R = 0.0821 L atm/K mol Atm= 4.15 x 10 -4 atm. T= 305K Concentration= (4.15 x 10 -4 atm) ÷ (0.0821 L atm/K molx 305K)= 0.000415 ÷ 25.0405= 1.657315x 10 -5 mol/L Concentration = 0.00001657315 Molar mass = Mass/ Concentration x 1/Volume (L) 0.608g/ 1.657315x 10 -5 mol/L x 1/(25÷1000) (36685.8443) x 40= 1.467x 10 6 g/Mol Molar mass= 1.467x 10 6 g/Mol 9. What is the predicted van't Hoff factor for NaF ? Ideally, the van't Hoff factor (i) for NaF is "2" The ideal values for the ionization of sodium fluoride is 2 since there is one Na + and one F - for each NaF that is dissolved 1 + 1 = 2
Jenele McLean 5

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