1. a 4.0 g mass of NaOH is dissolved in 5 mL of water. determine the aproximate

Question

1. a 4.0 g mass of NaOH is dissolved in 5 mL of water. determine the aproximate molar concentration of the NaOH.

2. refer to question 1 , if 4 Ml of the determined concentration is diluted to 500 mL, what is the concentration of this new batch of NaOH solution?

3.your lab partner and yourself arrive to lab one day and find an unknown concentration of NaOH solution on the top of te bench. out of curiosity, you ask your lab instructor if you can determine the concentration by performing a titration analysis using 30 mL of .25 M HCl. in the end, you titrated 35 mL of the unknown NaOH. what is the concentration of the unknown NaOH?

4. what is the molarity of a solution that contains 1.85 moles of H2SO4 in 2.50 L of solution?

5. what is the molarity if a solution prepared by dissolving 25.0 g of HCl (g) in enough water to make 150.0 mL of solution?

6. your lab instructor instructed the class to prepare 400 mL of 1.2 M solution of HNO3, how many grams of HNO3 is required to make the solution?

7. how many moles of CuSO4 are contained in 300 mL of a 0.40 M CuSO4 solution?

8. battery acid is generally 5 M H2SO4 . how many grams of H2SO4 are ub 170 mL of this solution?

Explanation / Answer

molarity of a substance in a solution = (number of mols of substance in the solution) / (volume of soltion in litres)

1)) molar mass of NaOH = 23+16+1 = 40g

4 g NaOH = 4/40 mols = 0.1 mol

volume of the soln = 5 ml = 0.005 ltr

molarity = 0.1/0.005 = 20M

2)) From question 1 we know that the molarity of the original solution is 20.

4 ml = 0.004 ltr

so 4 ml of this soln will contain 0.004 X 20 = 0.08 mols of NaOH.

volm of the new soln = 500 ml = 0.5 ltr

molarity of the new soln = 0.08/0.5 = 0.16M

3)) HCl + NaOH = NaCl + H2O

1 mol of HCl requires 1 mol of NaOH

no of mols of HCl = molarity X volume in ltr = 0.25 X 0.03 = 0.0075

1 mol of HCl requires 1 mol of NaOH.

So 0.0075 mols require 0.0075 mol NaOH.

So 35 ml of soln contains 0.0075 mol NaOH.

molarity = 0.0075/0.035 = 0.214M

4)) 1.85 mols of H2SO4 is present in 2.5 ltr

molarity = 1.85/2.5 = 0.74M

5)) molar mss of HCl = 1+35.5 = 36.5 g

25 g HCl = 25/36.5 = 0.685 mol

volume = 150 ml = 0.015 ltr

molarity = 0.685/0.015 = 45.66M

6)) volue of HNO3 soln = 400 ml = 0.4 ltr

molarity = 1.2M

no of mols of HNO3 = 1.2 X 0.4 = 0.48 mols

molar mass of HNO3 = 1 + 14 + 3 X 16 = 63 g

1 mol HNO3 weighs 63 g

0.48 mols weighs 0.48 X 63 = 30.24g

So 30.24 g of HNO3 is required.

7))

molarity of CuSO4 soln = 0.4M

volime of CuSO4 soln = 300 ml = 0.3 ltr

no of mols of CuSO4 = molarity X volume = 0.4 X 0.3 = 0.12M

8))

molar mass of H2SO4 = 2 X 1 + 32 + 4 X 16 = 98g

molarity of the H2SO4 soln = 5

volume of the H2SO4 soln = 170 ml =0.17 ltr

no of mols H2SO4 in 170 ml = molarity X volume in ltr = 0.17 X 5 = 0.85

mass of 0.85 mols H2SO4 = molar mass of H2SO4 X 0.85 = 83.3g

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.