4. a. What is the molality of a solution, if 250.0 g of NaCl is dissolved into 4

Question

4. a. What is the molality of a solution, if 250.0 g of NaCl is dissolved into 475.0 g of

solvent (water)?

What is the molarity of the solution in the previous question, if the solution’s density is 1.083 g/ml.

5. Ammonium nitrate, NH4NO2, decomposes in solution, as shown here.

NH4NO2 (aq) ® N2(g) + 2H2O (1)

The concentration of NH4NO2 ion at the beginning of an experiment was 0.650 M. After 3.50 hr, it was 0.288 M. What is the rate constant value?

6. The decomposition of N2O5 to NO2 and O2 is first order, with a rate constant of

4.80 x 10-4/s at 45° C.

a. If the initial concentration is 2.88 x 10-2 M, what is the concentration after

950.0 s?

b. How long would it take for the concentration of N2O5to decrease to 1.50 x 10-2 M from its initial value of 2.88 x 10-2 M?

7. In a first-order reaction, the t 1/2 was found to be 8.5 hours. What is the value for the rate constant?

Explanation / Answer

4) Molality= moles of solute ( NaCl)/ kg of solvent

Moles= mass/Molecular weight and molecular weight of NaCl =58.5

Moles of NaCl = 250/58.5=4.273

Molality= 4.273/0.475 kg of water =8.99 molal

Molarity = moles of solute/ L of solution

Mass of the mixture = 475+250= 725gm

Volume of the solution = 725/1.083 ml =669.4ml =0.6694 L

Molarity= 4.273/0.6694=6.383M

5)

The decomposition of NH4NO2 is 1st order whose equation is given by

-ln (1-XA)= Kt

XA= 1-CA/CAO, CAO =initial concentration = 0.650, CA= Concentration at any time t= 0.288M

XA= 1-0.288/0.650=0.5569

-ln(0.5569)= K*3.5 ( K is rate constant)

K= 0.167/hr

6.

-ln (1-XA)= Kt

XA= 1-CA/CAO, CAO =initial concentration =2.88*10-2 M, CA= Concentration at any time t= 950 seconds

-ln(1-XA)= 4.8*10-4*950=0.456

1-XA= 0.6338

XA= 1-0.6338=0.3662

CA/CAO= 0.3662*2.88*10-2 =0.010547 M

b) from

-ln(1- 1.5/2.88)= 4.8*10-4*t

Time, T =1532.7 seconds

7. for 1st order reaction, half life = 0.693/K

K= rate constant, K= 0.693/8.5 = 0.0815.hr

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