1. a. You set up a titration experiment in lab, in order to determine the molari

Question

1. a. You set up a titration experiment in lab, in order to determine the molarity of a nitric acid solution, HNO3 (aq). A flask containing 14.92 mL of a nitric acid solution was titrated with 0.135 M NaOH. 16.93 mL of the base were required to reach the endpoint.

What is the concentration of the nitric acid solution?

HNO3 (aq) + NaOH (aq) ightarrow NaNO3 (aq) + H2O (l)

molarity:_________________

b. Let’s separate the different data pieces of the HNO3 solution into solute, solvent and solution. Using the density of nitric acid solution (1.09 g/mL) and the molarity you calculated above and the volume used

Volume of original nitric acid solution: _______________

Grams & Moles of Nitric Acid? ___________________

Grams of Water? ______________________

Total Solution? _______________

In another titration, a student uses 1.023 M HCl solution to standardize a sodium hydroxide solution. Assuming 1.00 L of HCl solution and its density of 1.02 g/mL, calculate the mass percent and molality of the HCl solution.

mass percent: _____________

molality: ____________

2. Draw the expanded structure of ethanoic acid (acetic acid)

4. What is the solute in the vinegar solution?

Explanation / Answer

1. NaOH + HNO3 -------------------- NaNO3 + H2O

NaOH= 16.93 ml of 0.135M

number of moles of NaOH= 0.135M x 0.01693 L= 0.002286 moles

According to equation

1 mole of NaOH= 1 mole of HNO3

0.002286 moles of NaOH= 0.002286 moles of HNO3

number of moles of HNO3 = 0.002286 moles

volume of HNO3 = 14.92 ml = 0.01492L

Molarity of HNO3 = number of moles/volume in L = 0.002286/0.01492 = 0.153M

Molarity of HNO3 = 0.153M

Concentration of HNO3 = 0.153M

2.Ethanoic acid = CH3COOH = H3C-COOH

4.In the vinegar solution , acetice acid as a solute.

In general 10% acetice solution is called as Vinegar.

3. Molarity of HCl= 1.023M

volume of HCl = 1L = 1000 ml

density of the solution = 1.02 g/ml

mass of HCl= 1.02 gram

Mass of solution = 1.02 x 1000 = 1020 grams

mass percent = mass of solute/mass of solutionx 100

mass percent = 1.02/1020 x100 = 0.1%

mass percent = 0.1%

Molarity = 1.023M

density = 1.02 g/ml

Molality = 1000xM/(1000xd)-(MxMW)

Molality = 1000x 1.023/(1000x1.02)-(1.023x36.5)

Molality = 1023/982.66

Molality = 1.04m

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