(17) [10pts; Show work for credit; Mind the units] An aqueous solution has an os
ID: 702219 • Letter: #
Question
(17) [10pts; Show work for credit; Mind the units] An aqueous solution has an osmotic pressure of 12.5 mmHg at 300K. The solution was made by dissolving 20.0 mg (milligram) of solute in a small amount of solvent and then bringing the final volume of the solution to 5.00mL (a) [1pt] What is the solvent of the solution? (b) [3pts] Calculate the molarity of the solution? (c) [3pts] Calculate the number of moles of solute in the solution? (d) [3pts] Calculate the molar mass of the solute? (18) [20pts; Show work for credit; Mind the units] The concentration-time data for 2NO(g)- 2NO(g) +O2(g) at 300°C are tabulated and plotted as follows: In INO2 INO2I 8.00 x 10-3 6.58 x 10 3 559 x 10 4.85 x 10-3 4.29 x 10-3 3.48 x 10 2.93 x 10-3 2.53 x 103 Time (s) 0 50 100 150 200 300 400 500 -4.828 -5.024 5.187 -5.329 5.451 5.661 -5.833 -5.980 1/IN021 125 152 179 206 233 287 341 395 400 -4.90 300 5.30 200 -5.70 100 -6.10 0 100 200 300 400 500 0 100 200 300 400 500 Time (s) Time (s) (a) [2pts] Write an expression for the reaction rate specific for the reaction given above based on concentration change. Be as specific as possible. (b) [2pts] Based on the plots given above, what is the overall reaction order? Justify your answer. (c) [2pts] Write the rate law for the reaction based on your answer in (b) (d) [2pts] Which part/term of your rate law expression in (c) is temperature-dependent? (e) [4pts] Calculate the rate constant at 300°C? (f) [4pts] Calculate the concentration of NO2 at t-20.0min? (9) [4pts] Calculate ti2 when [NO2lo is 6.00x10 M?
Explanation / Answer
Ans 17
Part a
Aqueous solutions contain water as a solvent
Part b
Osmotic pressure = M x R x T
12.5 mmHg x 1atm/760mmHg = M x 0.0821 L-atm/mol-K x 300K
M = 0.0006677 mol/L = 6.67 x 10^-4 mol/L
Part C
Moles of solute = molarity x volume
= 0.0006677 mol/L x 5 mL x 1L/1000 mL
= 3.338 x 10^-6 mol
Part d
Molar mass = mass/moles
= (20 mg x 1g/1000mg) / (3.338 x 10^-6 mol)
= 5991.61 g/mol
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