He started the experiment by measuring the mass and freezing point of the pure solvent with minimal error. Instead of quickly finishing the experiment, he took a lunch break. After returning from lunch, he noted that the solvent had absorbed some imp atmosphere. He went ahead and performed the rest of the experiment anyway. Assuming a negligible (very minor) change in the mass of the solvent after absorbing impurities, which of the following choices contains only ca because of the experimental error, the measured AT will increase, the measured molality will increase, and the measured molar mass w additional error O because of the experimental error, the measured ATwill decrease, the measured molality will decrease, and the measured molar mass because of the experimental error, the measured ATwill decrease, the measured molality will increase, and the measured molar mass because of the experimental error, the measured ATwill decrease, the measured molality will increase, and the measured molar mas O because of the experimental error, the measured AT will increase, the measured molality will increase, and the measured molar mas

He started the experiment by measuring the mass and freezing point of the pure solvent with minimal error. Instead of quickly finishing the experiment, he took a lunch break. After returning from lunch, he noted that the solvent had absorbed some imp atmosphere. He went ahead and performed the rest of the experiment anyway. Assuming a negligible (very minor) change in the mass of the solvent after absorbing impurities, which of the following choices contains only ca because of the experimental error, the measured AT will increase, the measured molality will increase, and the measured molar mass w additional error O because of the experimental error, the measured ATwill decrease, the measured molality will decrease, and the measured molar mass because of the experimental error, the measured ATwill decrease, the measured molality will increase, and the measured molar mass because of the experimental error, the measured ATwill decrease, the measured molality will increase, and the measured molar mas O because of the experimental error, the measured AT will increase, the measured molality will increase, and the measured molar mas

Chemistry: An Atoms First Approach

2nd Edition

ISBN:9781305079243

Author:Steven S. Zumdahl, Susan A. Zumdahl

Publisher:Steven S. Zumdahl, Susan A. Zumdahl

Chapter16: Spontaneity, Entropy, And Free Energy

Section: Chapter Questions

Problem 114CP: Carbon tetrachloride (CCl4) and benzene (C6H6) form ideal solutions. Consider an equimolar solution...

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Carbon tetrachloride (CCl4) and benzene (C6H6) form ideal solutions. Consider an equimolar solution of CCl4 and C6H6 at 25C. The vapor above the solution is collected and condensed. Using the following data, determine the composition in mole fraction of the condensed vapor. Substance Gfo C6H6(l) 124.50 kJ/mol C6H6(g) 129.66 kJ/mol CCI4(l) 65.21 kJ/mol CCI4,(g) 60.59 kJ/mol
The weight fraction of methanol in an aqueous solution is 0.64. The mole fraction of methanol (x) satisfies a. X < 0.5 b. 0.5 < XM < 0.64 c. xu = 0.5 d. Xu ~ 0.5
The partial molar volumes of propanone (acetone) and trichloromethane (chloroform) in a mixture in which the mole fraction of CHCl3 is 0.4693 are 74.166 cm3 mol−1 and 80.235 cm3 mol−1, respectively. What is the volume of a solution of mass 1.000 kg?
The partial molar volumes of acetone (MW = 58.08 g mol−1) and chloroform (MW = 119.37 g mol−1), respectively, 74.166 cm3 mol−1 and 80.235 cm3 mol−1 for a solution in which the mole fraction of chloroform is 0.4693. What is the volume of 1.000 kg of this solution?
The partial molar volumes of propanone and trichloromethane in a mixture in which the mole fraction of CHCI3 is 0.4693 are 74.166 cm3 mol-1, and 80.235 cm3 mol-1, respectively. What is the volume of a solution of total mass 1.000 kg?
At 25 °C, the mass density of a 50 per cent by mass ethanol–water solution is 0.914 g cm−3. Given that the partial molar volume of water in the solution is 17.4 cm3 mol−1, calculate the partial molar volume of the ethanol.
The partial molar volumes of two liquids A and B in a mixture in which the mole fraction of A is 0.3713 are 188.2 cm3 /mol and 176.14 cm3 /mol, respectively. The molar masses of A and B are 241.1 g/mol and 198.2 g /mol . What is the volume of a solution of mass 1.000 kg?
In class, we derived the expression:ln XA =(ΔvH/R) ((1/T)-(1/T°))where A represents the solvent (and B would represent the solute, such that XA + XB = 1).Using the approximations given below, derive the expression:∆Tb = XB (RT°2/ΔvH)where ∆Tb is the change in boiling point between the solution and the pure liquid, T − T°.Approximations:ln(1 − X) ≈ −XTT° ≈ T°2Note: this derived expression is often simplified in General Chemistry as ∆Tb = Kbm where Kb is the boiling point elevation constant, and m is the molality.
the partial molar volumes of acetone and chloroform in a mixture in which the mole fraction of CHCl3 is 0.4693 are 74.166cm3/mol and 80.235 cm3/mol, respectively. What is the volume of a solution of mass 1.000 kg?
Illustrate the calculation of the ppm concentration of a known amount of vaporised liquid in 1 L of closed vessel?
At 20 °C, the mass density of a 20 per cent by mass ethanol–water solution is 968.7 kg m−3. Given that the partial molar volume of ethanol in the solution is 52.2 cm3 mol−1, calculate the partial molar volume of the water.
Determine the mole fraction of ethanol in a 10% by volume aqueous ethanol solution. Given: Methanol = 46 g mol-1, M(H2O) = 18 g mol-1 , density of ethanol = 0.789 g cm-3 and density of (H2O) = 1 g cm-3.