Solutions and molarity
This presentation is the property of its rightful owner.
Sponsored Links
1 / 7

Solutions and Molarity PowerPoint PPT Presentation


  • 82 Views
  • Uploaded on
  • Presentation posted in: General

Solutions and Molarity. 1.13.8 L of CO 2 @ STP is dissolved in 250.0 mL of water. What is the molarity of the resulting solution?. 13.8 L CO 2. x 1 mol CO 2. 22.4 L. Molarity. =. 0.250 L. = 2.46 M.

Download Presentation

Solutions and Molarity

An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Presentation Transcript


Solutions and molarity

Solutions and Molarity


Solutions and molarity

1.13.8 L of CO2 @ STP is dissolved in 250.0 mL of water. What is the molarity of the resulting solution?

13.8 L CO2

x 1 mol CO2

22.4 L

Molarity

=

0.250 L

= 2.46 M


Solutions and molarity

  • If a 1.89 M solution is made by dissolving chlorine gas in 600.0 mL of water, what volume of Cl2 gas was dissolved?

x 22.4 L

= 25.4 L of Cl2 gas

0.600 L

x 1.89 mol

1 mol

1 L


Solutions and molarity

3.Calculate the molarity of a solution when 2.96 x 1024 FU’s of CaF2 are dissolved in 0.500 L of water?

2.96 x 1024 FU

x 1 mol

6.02 x 1023 FU

Molarity =

= 9.83 M

0.500 L


Solutions and molarity

4.How many molecules are dissolved in 1.00 L of water if the molarity of the H2SO4 solution is 0.2 M?

x 0.2 mol

x 6.02 x 1023 molecules

1.00 L

= 1 x1023 molecules

1 mol

1L


Solutions and molarity

5.Excess sodium hydroxide solution is added to 80.0 mL of 0.336 M ZnCl2, calculate the mass of zinc hydroxide that will precipitate.

2 NaOH (aq)+ZnCl2(aq)→ Zn(OH)2(s)+ 2NaCl (aq)

0.0800 L

? g

x 1 mole Zn(OH)2

0.0800 L ZnCl2

x 0.336 mole

x 99.4 g

1 L

1 mole ZnCl2

1 mol

= 2.67 g Zn(OH)2


Solutions and molarity

6.Calculate the volume of 0.100 M HCl solution that is required to completely neutralize 250.0 mL of 0.200 M Ba(OH)2 solution.

2HCl(aq) + Ba(OH)2(aq) → BaCl2(aq) + 2H2O (l)

? L

0.2500 L

x 2 mole HCl

0.2500L Ba(OH)2

x 0.200 mole

x 1 L

1 L

1 mole Ba(OH)2

0.100 mol

= 1.00 L


  • Login