1 / 11

Molar Concentrations

Molar Concentrations. Molarity is the number of moles of solute that can dissolve in 1 L of solution. Molar concentration (mol/L) =. Amount of solute (mol) Volume of solution (L). n. V. C.

delta
Télécharger la présentation

Molar Concentrations

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. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Molar Concentrations

  2. Molarity is the number of moles of solute that can dissolve in 1 L of solution. Molar concentration (mol/L) = Amount of solute (mol) Volume of solution (L) n V C

  3. What is the concentration, in mol/L, of a solution formed by dissolving 28.0g of calcium chloride in enough water to make 225 mL of solution? Step 1: Find the number of moles of calcium chloride using n=m/M n= = = 0.252 mol CaCl2 m M 28.0 110.98

  4. Step 2: Use c=n/V to calculate the molar concentration c= = = 1.12 mol/L n V Notice the volume has been converted from mL to L 0.252 0.225 L

  5. How many grams of sodium nitrate would be needed to make 425 mL of 6.00 mol/L solution? Step 1: Find the number of moles using n= cV n= (6.00 mol/L)(0.425L) = 2.55 mol of sodium nitrate Step 2: Calculate how many grams is needed using m = nM m = (2.55 mol) (85.00 g/mol) = 217 g of NaNO3

  6. What final solution volume would be required to prepare A 0.100 mol/L solution of AgNO3(aq) if 1.20 g of the solid salt will be used? MMAgNO3 = 169.9 g/mol C = 0.100 mol/L mAgNO3 = 1.20 g V = ? nAgNO3 = m/MM = 1.20 g 169.9g/mol = 0.00706 mol V = n/C = 0.00706 mol/0.100 mol/L = 0.0706 L

  7. Conversion of % to mol/L Calculate the concentration, in mol/L, of a 96% solution of sulphuric acid which has a density of 1.84 g/mL Step 1. Calculate the number of moles of sulphuric acid. n= m/M = = 0.98 mol of H2SO4 96g 98.0 g/mol

  8. Step 2: Find the volume of the solution using the mass and the density. Recall: V= m/D = = 54.3 mL = 0.0543 L of solution m 100 g 1.84 g/mL D V

  9. Step 3: Calculate the concentration using c= n/V c = n/V = = 18 mol/L The concentration is 18 mol/L 0.98 mol of H2SO4 0.0543L

  10. Examples Eg1) A Solution of sodium chloride in water contains 14.0 g of NaCl dissolved in 250 mL of solution. What is the molarity of the solution? What do we need to find? Given? MMNaCl = 58.44 g/mol mNaCl = 14.0 g Vsolution = 250 mL C = ? C = n/V n = m/MM = 14.0 g/58.44 g/mol = 0.240 mol NaCl C = n/V = 0.240 mol/0.250 L = 0.960 mol/L

  11. Examples Eg 2) What mass of sucrose, C12H22O11, is dissolved in 50.0 mL of a 0.400 mol•L-1 solution of sucrose in water? Given, MMsucrose = 342.3 g/mol C = 0.400 mol/L V = 50.0 mL = 0.0500 L ! nsucrose= C x V = 0.400 mol/L x 0.0500 L = 0.0200 mol sucrose msucrose = n x MM = 0.0200 mol x 342.3 g/mol = 6.85 g of sucrose

More Related