Monday, September 19, 2011

AP Chem blog 6 due mon 9/26 at 9:00 am

Perform the following serial dilutions:

1.75 L of 0.33 M hydrochloric acid from a stock solution of 10 M hydrochloric acid

250 mL of .01 M acetic acid from a solution of 1.5 M acetic acid solution.

Use dimensional analysis to solve!

14 comments:

  1. 1.75L HCl × (.33 mol HCl)/(1 L solution)= .578 mol HCl

    .578 mol HCl × (1 L solution)/(10 mol HCl)= .0578 L of 10 M HCl solution

    First add .05775 L of 10 M HCl solution into a graduated cylinder, then add distilled water until the 1.75 mark is reached.

    .250L C2H4O2 × (.01 mol C2H402)/(1 L sol)= .00250 mol C2H4O2

    .00250 mol C2H4O2 × (1 L solution)/(1.5 mol C2H402 )= .00167 L of 1.5 M C2H4O2 solution

    First add .00167 L of 1.5 M C2H4O2 solution into a graduated cylinder, then add distilled water until the 250 mL mark is reached.

    ReplyDelete
  2. a) stock: 10M HCL
    1.75 L HCl * (.33 mol /1 L) = .5775 mol HCL

    .5775 mol HCL * (1L HCL / 10 mol) = .05775 L HCL
    Add .05775 L of the 10 M solution into the volumetric flask and then add water until the 1.75 L mark. add water slowly and stir constantly.

    b) 250 mL * (1 L / 1000 mL) = .250 L CH3COOH
    .250 L CH3COOH * (.1 mol/ 1L) = .025 mol CH3COOH
    .025 mol CH3COOH * (1.5 mol/ 1 L) = .0375 L CH3COOH

    add .0375 L CH3COOH stock solution into the volumetric flask, and then add water until the .250 L line. add water slowly and stir constantly.

    ReplyDelete
  3. 1.75 L HCl × (.33 mol HCl / 1 L sol) x (1 L solution / 10 mol HCl) = .0578 L of 10 M HCl sol

    Measure out .0578 L, or 57.8 mL of 10 M HCl solution into a graduated cylinder. Then, pour water until the 1.75 L mark.

    .250 L acetic acid × (.01 mol acetic acid / 1 L sol) × (1 L sol /1.5 mol acetic acid )= .00167 L of 1.5 M acetic acid solution

    Measure out .00167 L, or 1.67 mL of 10 M HCl solution into a graduated cylinder. Then, pour water until the 250 mL, or .250 L mark.

    ReplyDelete
  4. a)(1.75)(.33)=.5775
    (.5775)/10=.05775
    Put the .05775 into a graduated cylinder. Pour water til the 1.75L mark.
    b)(250)(.01)/1000=.0025
    (.0025)/1.5=.00167
    Put .00167L of the 10M into a graduated cylinder. then pour to the 250 ml

    ReplyDelete
  5. a) Diluting 10 M HCl solution to make 0.33 M HCl solution of 1.75 L in volume
    1.75 L HCl x (0.33 mol HCl/1 L solution) = 0.5775 mol HCl
    0.5775 mol HCl x (1 L solution/10 mol HCl) = 0.05775 L solution of HCl
    Add 0.05775 L of the stock solution into a graduated cylinder, add water slowly and each time you add water stir the solution. Repeat this process of adding water and stirring until the 1.75 L mark has been reached.

    b) Diluting 1.5 M CH3COOH solution to make 0.01 M CH3COOH solution of 250 mL in volume
    250 mL CH3COOH solution x (1 L CH3COOH solution/1000 mL CH3COOH solution) = 0.250 L CH3COOH solution
    0.250 L CH3COOH x (0.01 mol CH3COOH/1 L solution) = 0.00250 mol CH3COOH
    0.00250 mol CH3COOH x (1 L solution/1.5 mol CH3COOH) = 0.00167 L solution
    Add 0.00167 L of the stock solution into a graduated cylinder, add water slowly and each time you add water stir the solution. Repeat this process of adding water and stirring until the 0.250 L or 250 mL mark has been reached.

    ReplyDelete
  6. a) 1.75LHClx(.33molHCl/1L)x(1L/10molHCl)x(1000mL/1L)= 57.8mLHCl

    Pour 57.8mLof the stock solution, then add water until it reaches 1.75L.

    b) 250mLx(.o1molHC2H3O2/1000mL)x(1000mL/1.5mol)= 1.67mLHC2H3O2

    Pour 1.67mL of the stock solution then add water until it reaches the 250mL mark.

    ReplyDelete
  7. a.) 1.75 L HCL x (.33 moles HCl/1L) x (1L/10 moles HCl)=.0578 L= 57.8 mL HCL
    Pour 57.8 mL of HCL from the stock solute to a graduated cylinder then add water until it reaches the 1.75 L mark

    b.) .250 L C2H3O2 x (.01 moles C2H3O2/1L) x (1L/1.5 moles C2H3O2)= .001666 L= 1.67 mL C2H3O2

    Pour 1.67 mL C2H3O2 from the stock solute to a graduated cylinder then add water until it reaches the 250 mL mark

    ReplyDelete
  8. A) 1.75 L HCl x (0.33mol HCl/1L sol) x (1L sol/10mol HCl = 0.578 mol HCl
    Pour 0.578 mol HCl of the 10M solution into a graduated cylinder and then add water until the the 1.75 L mark

    B) 250 mL x (1L/1000mL) x (.1mol/1L) x (1.5mol/1L) = 0.0375 L CH3COOH
    pour 0.0375 L of CH3COOH into a graduated cylinder, add water until the .250 L mark. While doing this you should slowly stir the solution inside the flask.

    ReplyDelete
  9. A)(1.75 L HCl) X (0.33 mol/1L) X (1L/10 mol)
    =0.0578 L HCl = 57.8 mL

    Obtain a 175 mL volumetric flask, pour 57.8 mL of stock solution and add water until it reached the 175 ml mark.

    B)(0.25 L C2H3O2) X (0.01 mol/1L) X (1L/1.5 mol)
    =1.66X 10^-3 L = 1.67 mL

    Obtain a 250 mL volumetric flask, pour 1.67 mL of the stock solution and add water until it reached the 250 mL mark.

    ReplyDelete
  10. 1.75 L HCl x (0.33 mol HCl/1 L solution)= 0.5775 mol HCl
    0.5775 mol HCl x (1 L solution/10 mol HCl)= 0.0578 L of 10 M HCl solution

    So, you pour 0.0578 L of 10 M HCl solution into a graduated cylinder and then add distilled water to it until the solution reaches the 1.75 L mark.

    0.250 L HC2H3O2 x (0.01 mol HC2H3O2/1 L solution) x (1 L solution/1.5 mol HC2H3O2)= (1.67 x 10^-3) L of 1.5 M HC2H3O2 solution

    So you pour (1.67 x 10^-3) L of 1.5 M HC2H3O2 solution into a graduated cylinder, and then add distilled water to it until the solution reaches Then, pour water until the solution reaches the 0.250 L mark.

    ReplyDelete
  11. A)1.75LHClx(0.33 mol HCl/1 L solution)=0.5775molHCl
    0.5775 molHCl x(1 L solution/10 mol HCl)=0.05775
    Put 0.05775 L of the stock solution in a graduated cylinder then add water 1.75 L mark.

    B).250 L C2H3O2 x(.01 moles C2H3O2/1L)x(1L/1.5 moles C2H3O2)=.001666 L=1.67mLC2H3O2
    Pour (1.67 x 10^-3)L 1.5 M HC2H3O2 solution into a graduated cylinder then add water til the solution reaches 0.250 L mark.

    ReplyDelete
  12. a) M1V1=M2V2
    (1.75 L)(.33 M)=(10 M)(V2)
    V2= .05775 L

    b) M1V1=M2V2
    (.01 M)(.25 L)=(1.5 M)(V2)
    V2= 1.67 *10 ^-3

    ReplyDelete
  13. a)1.75 L HCl x 0.33 mol HCl/1 L solution= 0.5775 mol HCl
    0.5775 mol HCl x 1 L solution/10 mol HCl= 0.0578 L HCl
    You pour your 0.0578 L of HCL solution into a graduated cylinder then add distilled water to it until the solution reaches the 1.75 L mark.Then you will have a 10 M HCL solution

    b)0.250 L HC2H3O2 x 0.01 mol HC2H3O2/1 L solution x 1 L solution/1.5 mol HC2H3O2= .00167 L C2H3O2
    FIrst You pour 0.00167 L of 1.5 M HC2H3O2 solution into a graduated cylinder then add distilled water to to the solution until it reaches the 0.250 L mark.

    ReplyDelete
  14. Pour 0.0578 L OF HCL SOLUTION add water until 1.75L
    Pour 0.00167L of 1.5M HC2H3O2 into a graduated cylinder and fill water until 0.25 L

    ReplyDelete