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ID: 575947 • Letter: Y
Question
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1) What mass (in g) of NH3 must be dissolved in 475 g of methanol to make a 0.250 m solution?
2) How many moles of KF are contained in 347 g of water in a 0.175 m KF solution?
3) Calculate the molality of a solution that is prepared by mixing 25.5 mL of CH3OH (d = 0.792 g/mL) and 387 mL of CH3CH2CH2OH (d = 0.811 g/mL).
4) What is the mol fraction of ethanol, C2H5OH, in a solution of 75.3g of ethanol and 525g of water?
5) Determine the freezing point of a solution that contains 78.8 g of naphthalene (C10H8, molar mass = 128.16 g/mol) dissolved in 722 mL of benzene (d = 0.877 g/mL). Pure benzene has a melting point of 5.50°C and a freezing point depression constant of 4.90°C/m.
6) Determine the boiling point of a solution that contains 78.8 g of naphthalene (C10H8, molar mass = 128.16 g/mol) dissolved in 722 mL of benzene (d = 0.877 g/mL). Pure benzene has a boiling point of 80.1°C and a boiling point elevation constant of 2.53°C/m.
7) Determine the rate law and the value of k for the following reaction using the data provided.
S2O82(aq) + 3 I(aq) 2 SO42(g) + I3(aq) [S2O82]i (M) [I]i (M) Initial Rate
0.30 0.42 4.54
0.44 0.42 6.65
0.44 0.21 3.33
8) Determine the rate law and the value of k for the following reaction using the data provided.
NO2(g) + O3(g) NO3(g) + O2(g) [NO2]i (M) [O3]i (M) Initial Rate
0.10 0.33 1.42
0.10 0.66 2.84
0.25 0.66 7.10
9) A reaction is found to have an activation energy of 108 kJ/mol. If the rate constant for this reaction is 4.60 × 10-6 s-1 at 275 K, what is the rate constant at 366 K?
10) Given the following proposed mechanism, predict the rate law for the overall reaction.
2NO2 + Cl2 2NO2Cl (overall reaction)
Mechanism
NO2 + Cl2 NO2Cl + Cl slow
NO2 + Cl NO2Cl fast
A) Rate = k[NO2][Cl2]
B) Rate = k[NO2]2[Cl2]2
C) Rate = k[NO2][Cl]
D) Rate = k[NO2Cl][Cl]2
E) Rate = k[NO2Cl]2
Explanation / Answer
Ans 1 :
Molality = no. of moles of NH3 / mass of methanol in kg
0.250 = n / 0.475
n = 0.11875 mol
Mass of NH3 = 0.11875 x molar mass
= 0.11875 x 17.031
= 2.02 grams
Ans 2 :
m = no. of moles / mass of solvent in kg
0.175 = n / 0.347
n = 0.060725 mol
Ans 3 :
Mass of CH3OH = volume x density
= 25.5 x 0.792
= 20.196
No. of moles of CH3OH = 20.196 / molar mass
= 20.196 / 32.04
= 0.63 mol
Mass of propanol = 387 x 0.811
= 313.857 g
No. of moles of propanol = 313.857 / 60.09
= 5.22 mol
Molality = n / mass of solvent in kg
m = 0.63 / 0.314
m = 2.00 m
Ans 4 :
Number of moles of ethanol = given mass / molar mass
= 75.3 / 46.07
= 1.634 mol
Number of moles of water = 525 / 18.01528
= 29.141 mol
Total number of moles = 1.634 + 29.141
= 30.775 mol
Mole fraction of ethanol = no. of moles of ethanol / total number of moles
= 1.634 / 30.775
= 0.053
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