How do you prepare 500 ml of a 1.77M H2SO4 solution from an 18.0 M H2SO4 stock solution?

Molarity Problems Worksheet  

M=nV  n= # moles  

V must be in liters (change if necessary)  

1. What is the molarity of a 0.30 liter solution containing 0.50 moles of NaCl?  

1.7M

2. Calculate the molarity of 0.289 moles of FeCl3 dissolved in 120 ml of solution?  

2.41 M

3. If a 0.075 liter solution contains 0.0877 moles of CuCO4, what is the molarity?  

1.2M

4. How many moles of NaCl are present in 600. ml a 1.55 M NaCl solution?  

.930 moles

5. How many moles of H2SO4 are present in 1.63 liters of a 0.954 M solution?  

1.56 molse

6. How many liters of solution are needed to make a 1.66 M solution containing 2.11 moles of KMnO4?  

1.27 g

7. What volume of a 0.25 M solution can be made using 0.55 moles of Ca(OH)2?  

2.2 L

For all of the problems below you will need to do a mole-mass conversion. Each problem will involve two steps.  

8. What is the molarity in 650. ml of solution containing 63 grams of NaCl?  

1.7 M

9. How many grams of Ca(OH)2 are needed to produce 500. ml of 1.66 M Ca(OH)2 solution?  

61.5 g

10. What volume of a 0.88 M solution can be made using 130. grams of FeCl2?  

1.2 L

Dilution Problems Worksheet  

1. How do you prepare a 250.-ml of a 2.35 M HF dilution from a 15.0 M stock solution?  

39.2 mL

2. If 455-ml of 6.0 M HNO3 is used to make a 2.5 L dilution, what is the molarity of the dilution?  

1.1 M

3. If 65.5 ml of HCl stock solution is used to make 450.-ml of a 0.675 M HCl dilution, what is the molarity of the stock solution?  

4.64 M

4. How do you prepare 500.-ml of a 1.77 M H2SO4 dilution from an 18.0 M H2SO4 stock solution?  

Take 49.2-ml of 18.0 M H2SO4 stock solution and pour it into a 500-ml volumetric flask. Fill to the 500-ml line with distilled water to make 1.77M H2SO4 solution.

Extra Molarity Problems for Practice  

1. How many moles of LiF would be required to produce a 2.5 M solution with a volume of 1.5 L?  

3.75 M

2. How many moles of Sr(NO3)2 would be used in the preparation of 2.50 L of a 3.5 M solution?  

8.75 M

3. What is the molarity of a 500-ml solution containing 249 g of KI?  

3.00 M

4. How many grams of CaCl2 would be required to produce a 3.5 M solution with a volume of 2.0 L?  

777 g

Explanation:

I've done this before

  the   balanced   equation   for   reaction of   ca metal   with   o2 to   produce   cao   is as below

2 ca+ o2→ 2 cao2

  explanation

the reaction is   balanced     since the   number of atoms   in reactant side is   equal to   number of atoms   in   product side.   for   example   they   are 2   atom of   ca in   reactant   side   and   2   atoms of ca   in   the product side.

  the   process   is a   meet the   requirement   of   redox   reaction because :     calcium   is   oxidized   from   oxidation state 0   to 2+   while   oxygen   is reduced from   oxidation   state   0   to 2-

a electronegativity decreases beacuse the size of nucleus is bigger.