Question

1. A solution is prepared by dissolving 20.2 mL CH,OH in 100.0 mL HO at 25 degrees Celsius. The final volume of the solution is 118.0 mL. The densities of CHOH and H,O are 0.782 g/mL and 1.00 g/mL, respectively. For this solution, calculate molality, molarity, mole fraction of each component and % by mass of each component. 2. A solution contains a mixture of pentane and hexane at room temperature. The solution has a vapor pressure of 258 torr. The vapor pressure of pure pentane is 425 torr and the vapor pressure of pure hexane is 151 torr at room temperature. What is the mole fraction of each component in the mixture? 3. Consider the reaction: CHCI, (g) + Cl, (g) → CCI, (g) + HCI (g). The initial rate of the reaction is measured at several different concentrations of the reactants with the results found in the table. Write the rate equation for the reaction after determining the order with respect to each reactant and calculate the rate constant, k. Trial (CHCI], [CL].M Initial Rate, M/sec M 1 0.010 20.020 3 0.020 4 0.040 0,010 0,0035 0.010 0.0069 0.020 0.0098 0.040 0.027

Answer 1

1) Firstly, the moles of CH3OH and H2O are calculated:

n CH3OH = 20.2 mL * (0.782 g / 1 mL) * (1 mol / 32 g) = 0.5 mol

n H2O = 100 mL * (1 g / 1 mL) * (1 mol / 18 g) = 5.56 mol

Is calculated:

molality = n / Kg = 0.5 mol / 0.1 Kg = 5 m

molarity = n / L = 0.5 mol / 0.118 L = 4.24 M

X CH3OH = n CH3OH / total n = 0.5 / 5.55 + 0.5 = 0.083

X H2O = 1 - 0.083 = 0.917

% mass CH3OH = 20.2 * 0.782 * 100 / (20.2 * 0.782) + (100 * 1) = 13.64%

% mass H2O = 100 - 13.64 = 86.36%