Question

1.) You have a solution that is 18.5% (v/v) methyl alcohol. If the bottle contains 1.27 L of solution, what is the volume (V) in milliliters of methyl alcohol?

2.) A 6.00 % (m/v) NaCl solution contains 26.8 g of NaCl. What is the total volume (V) of the solution in milliliters?

3.) 25.0 mL of a 6.0 M HNO3 stock solution is diluted using water to 100 mL. How many moles of HNO3 are present in the dilute solution?

25.0 mL of a 6.0 M  stock solution is diluted using water to 100 mL. How many moles of  are present in the dilute solution?

1.5 moles

0.6 moles

6.0 moles

0.15 moles

Submit

4.) How many mL of a 14.0 M NH3 stock solution are needed to prepare 200 mL of a 4.20 M dilute NH3 solution?

a.)667 mL b.)60.0 mL c.)840 mL d.)0.060 mL

5.) The four following concentrated solutions are each diluted with water to form 200 mL of a dilute solution. Which solution, once diluted to 200 mL, will have the largest concentration?

The four following concentrated solutions are each diluted with water to form 200 mL of a dilute solution. Which solution, once diluted to 200 mL, will have the largest concentration?

20.0 mL of a 0.4 M NaOH solution

80.0 mL of a 0.2 M NaOH solution.

10.0 mL of a 0.5 M NaOH solution

100.0 mL of a 0.1 M NaOH solution

Answer 1

1- V/V % means volume of solute in 100 mL of solution.

Now given- Solution has methyl alcohol = 18.5% (v/v)

Total volume of solution = 1.27 L = 1.27 * 1000 = 1270 mL

Thus amount of methyl alcohol in it =  1270 mL * 18.5 / 100

= 234.95 mL

2-  6.00 % (m/v) NaCl solution means if the solution is 100 mL , then amount of NaCl in it = 6g

Or 6g NaCl is present in 100 mL solution

Then 26.8 g of NaCl will be present in = (100 mL * 26.8 g)/ 6 g

= 446.66 mL solution

3- Intitial Volume (V1) of HNO3 solution = 25.0 mL

Intitial concentration (M1) of HNO3 solution = 6.0 M

Now the solution is diluted by using 100 mL water

Thus

final volume (V2) of the solution becomes = 25 + 100 = 125 mL

The relation between volume and concentration is inversely proportional. i.e as the volume increases, the concentration decreases and vice versa.

Hence final concentration (M2) will be-

M1 * V1 = M2 * V2

M2 = M1 * V1 / V2

= 6.0 M* 25.0 mL / 125 mL

= 1.2‬ M

Now no. of moles = Concentration * volume

= 1.2 M * 125 mL

= 1.2 mol * 125 ml / 1000 ml

= 0.15‬ mols

4- Initial volume (V1) =?

Initial conc (M1) = 14.0 M

Final vol (V2) = 200 mL

Final conc (M2) = 4.20 M

We know M1V1 = M2V2

Thus V1 = M2V2 / M1

= 200 mL * 4.20 M / 14.0 M

= 60‬ mL

Thus 60 mL of a 14.0 M NH3 stock solution are needed to prepare 200 mL of a 4.20 M dilute NH3 solution

5- Largest concentration means highest no. of moles of solute. Now no. of moles in the following solutions are

1- 20.0 mL of a 0.4 M NaOH solution

No. of moles = conc * vol

= 0.4 mole * 20 ml/ 1000 ml

= 0.008‬ mols

2- 80.0 mL of a 0.2 M NaOH solution.

  No. of moles = conc * vol

= 0.2 mole * 80 ml/ 1000 ml

= 0.016‬‬ mols

3-   10.0 mL of a 0.5 M NaOH solution

  No. of moles = conc * vol

= 0.5 mole * 10 ml/ 1000 ml

= 0.005‬‬ mols

4- 100.0 mL of a 0.1 M NaOH solution

No. of moles = conc * vol

= 0.1 mole * 100 ml/ 1000 ml

= 0.01‬‬ mols

Thus the highest no of moles = 0.016‬‬ mols = 80.0 mL of a 0.2 M NaOH solution.

Hence it has the highest concentration after dilution.