When ethylene glycol, formula C_2H_6O_2 is used as antifreeze for automobiles it

Question

When ethylene glycol, formula C_2H_6O_2 is used as antifreeze for automobiles it is mixed with water to make a 30% (v/v) solution. Assume that the volumes are additive and that the solution behaves ideally. The following data were measured for the reaction BF_3(g) + NH_3 (g) rightarrow F_3 BNH_3 (g): What is the order of this reaction with respect to each reactant? What is the rate constant, with proper units, for this reaction? What is the rate law for this reaction? Cocaine is a weak organic base whose molecular formula is C_17H_23NO_4. At 15 degree C an aqueous solution of cocaine was found to have a pH of 8.53 and an osmotic pressure of 52.7 torr. What is the molarity of this solution? What is kb of cocaine? What concentration of NaOH solution would have the same pH as this solution of cocaine? Consider a mixture of 25.0 mL of 0.100 M HCl and 75.0 mL of 0.100 M acetic acid. Construct an ICE table Determine the pH of the solution Consider the reaction between hydrogen peroxide and iron in acidic solution Balance the reaction in acidic solution Determine the standard potential for this reaction Determine the potential for this reaction if the concentrations of hydrogen peroxide, iron (II), and iron (III) are all 0.10 M and the solution is pH = 4

Explanation / Answer

1. 30 g ethylene glycol in 100 ml water

molality of solution = 30/62.07 x 0.1 = 4.83 m

Volume of solution = 30 x 1.11 + 100 = 133.3 ml

molarity of solution = 0.483/0.1333 = 3.623 M

moles of ethylene glycol = 30/62.07 = 0.483 mol

moles of water = 100/18 = 5.555 mol

mole fraction of water = 5.555/(5.555 +0.483) = 0.92

a) freezing point = 0 - iKfm = 0 - 1 x 4.83 x 1.86 = - 8.984 oC

b) vapor pressure of solution = 0.92 x 17.54 = 16.137 torr

c) osmotic pressure = iMRT = 1 x 3.623 x 0.08205 x 298 = 88.585 atm at T = 25 oC

osmotic pressure = iMRT = 1 x 3.623 x 0.08205 x 273 = 81.154 atm at T = 0 oC

2.

a. let the rate law be,

rate = k[BF3]^x.[NH3]^y

x and y are order with respect to each reactant,

k is rate constant

From experiment 3 and 4, [NH3] is constant

0.1193/0.0682 = (0.35/0.2)^x

taking log,

x = 1

From experiment 1 and 2, [BF3] constant

0.1065/0.2130 = (0.125/0.250)^y

taking log

y = 1

Order with respect to both reactants is 1.

b. rate constant

k = 0.2130/(0.250)^2 = 3.408 M-1.s-1

c. rate law,

rate = k[BF3][NH3]

3. coccaine

a. osmotic pressure = 52.7 torr = 0.0693 atm

molarity = 0.0693/0.08205 x (273 + 15) = 0.003 M

b. pOH = 14 - pH = 5.47

[OH-] = 3.4 x 10^-6 M

Kb = (3.4 x 10^-6)^2/0.003 = 3.84 x 10^-9

c. NaOH concentration needed = 3.4 x 10^-6 M

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