VINCENT

1) What is the molarity of Na₂CO₃ when Na₂CO₃ is dissolve in water and the final volume is of the solution is 200 mL?

Molarity formula: M=n/V

M is Molarity

N is moles of solute

V is liters of solution

First convert 1.06 g Na₂CO₃ to moles since you need moles in liters not grams.

1mole of Na₂CO_{3 =} 106 g

1.06 g Na₂CO₃* (1mole of Na₂CO₃/106 g) = .01 moles of Na₂CO₃

Then covert 200mL of the solution to liters since you need liters not milliliters.

200ml*(1L/1000mL) = .2L

Then you use the molarity formula.

Molarity(M)= .01 moles of Na₂CO₃/.2L= .05M

Note to readers M is Moles/L

How can you prepare 200 mL of 0.5 solution of 0.5 M solution of MgCl₂ from a bottle of solid MgCl₂?

2) Another form of the Molarity formula: n=MV

First you convert 200 mL to liters since you need liters not milliliters.

200ml*(1L/1000mL) = .2L

Then you use the molarity formula

n=MV

n=(.5M)(.2L)                            Note to readers M is Moles/L

n =.1 mole MgCl₂

Since normally out of a bottle it will be grams not moles you convert moles to grams.

.1 mole MgCl₂*(95.3g/mole of MgCl2) = 95.3g

Now you have to explain how to prepare the solution.

First you measure 9.53 g of MgCl₂. Using a volumetric flask, dissolve the MgCl₂ with enough water (don’t fill all the way try using the less amount of water need). Lastly you fill the volumetric flask up to the 200 mL line.

3) Wang is going to explain this problem on Tuesday so this will be edited then.

Note to all, I made a program for the TI 84 so if you have one bring it so i can transfer it to your calculator. I’ll explain how to work it.