{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

CHM2046TEST1 - 1 CHM2046 TEST 1 Version A(Exam 3 Answers...

Info icon This preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
1 CHM2046 TEST 1 Version A (Exam 3) Answers underlined N A = 6.022×10 23 ; R = 8.314 J/mol.K = 0.08206 L.atm/mol.K; 760 torr = 760 mmHg = 1 atm 1) A 150.0 mL sample of an aqueous solution at 25°C contains 15.2 mg of an unknown nonelectrolyte compound. If the solution has an osmotic pressure of 8.44 torr, what is the molar mass of the unknown compound? a) 195 g/mol b) 223 g/mol c) 294 g/mol d) 341 g/mol e) 448 g/mol Use Π =MRT; solve for M; solve for molar mass M = Π /RT = (8.44/760)/ (0.08206 × 298) = 4.541 × 10 -4 mol/L mol = (4.541 × 10 -4 mol/L)(0.150 L)= 6.812 -5 mol molar mass = 0.0152 g/6.812 -5 mol = 223g/mol 2) Determine the vapor pressure of a solution at 25°C that contains 76.6 g of glucose (C 6 H 12 O 6 ) in 250.0 mL of water. The vapor pressure of pure water at 25°C is 23.8 torr. a) 23.1 torr b) 22.9 torr c) 70.8 torr d) 72.9 torr e) 7.29 torr Use P solvent = χ solvent P o solvent = 0.970 × 23.8 = 23.1torr 3) Determine H vap for a compound that has a measured vapor pressure of 24.3 torr at 273 K and 135 torr at 325 K. Use Clausius-Clapeyron eqn; solve for H vap = 325 1 273 1 3 . 24 135 ln R H vap ; 1.71 = H vap/R (5.83 × 10 -4 ) H vap = (1.71 × 8.314)/(5.83 × 10 -4 ) ÷ 1000 J
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon