### Molarity Problems#1 - 10

The equations I will use are:

M = moles of solute / liters of solution

and

MV = grams / molar mass <--- The volume here MUST be in liters.

Typically, the solution is for the molarity (M). However, sometimes it is not, so be aware of that. A teacher might teach problems where the molarity is calculated but ask for the volume on a test question.

Note: Make sure you pay close attention to multiply and divide. For example, look at answer #8. Note that the 58.443 is in the denominator on the right side and you generate the final answer by doing 0.200 times 0.100 times 58.443.

Problem #1: Sea water contains roughly 28.0 g of NaCl per liter. What is the molarity of sodium chloride in sea water?

Solution:

MV = grams / molar mass

(x) (1.00 L) = 28.0 g / 58.443 g mol¯1

x = 0.4790993 M

to three significant figures, 0.479 M

Problem #2: What is the molarity of 245.0 g of H2SO4 dissolved in 1.000 L of solution?

Solution:

MV = grams / molar mass

(x) (1.000 L) = 245.0 g / 98.0768 g mol¯1

x = 2.49804235 M

to four sig figs, 2.498 M

If the volume had been specified as 1.00 L (as it often is in problems like this), the answer would have been 2.50 M, NOT 2.5 M. You want three sig figs in the answer and 2.5 is only two SF.

Problem #3: What is the molarity of 5.30 g of Na2CO3 dissolved in 400.0 mL solution?

Solution:

MV = grams / molar mass

(x) (0.4000 L) = 5.30 g / 105.988 g mol¯1

0.12501415 M

x = 0.125 M (to three sig figs)

Problem #4: What is the molarity of 5.00 g of NaOH in 750.0 mL of solution?

Solution:

MV = grams / molar mass

(x) (0.7500 L) = 5.00 g / 39.9969 g mol¯1

(x) (0.7500 L) = 0.1250097 mol <--- threw in an extra step

x = 0.1666796 M

x = 0.167 M (to three SF)

Problem #5: How many moles of Na2CO3 are there in 10.0 L of 2.00 M solution?

Solution:

M = moles of solute / liters of solution

2.00 M = x / 10.0 L

x = 20.0 mol

Suppose the molarity was listed as 2.0 M (two sig figs). How to display the answer? Like this:

20. mol

Problem #6: How many moles of Na2CO3 are in 10.0 mL of a 2.0 M solution?

Solution:

M = moles of solute / liters of solution

2.0 M = x / 0.0100 L <--- note the conversion of mL to L

x = 0.020 mol

Problem #7: How many moles of NaCl are contained in 100.0 mL of a 0.200 M solution?

Solution:

0.200 M = x / 0.1000 L

x = 0.0200 mol

Problem #8: What weight (in grams) of NaCl would be contained in problem #7?

Solution:

(0.200 mol L¯1) (0.100 L) = x / 58.443 g mol¯1 <--- this is the full set up

x = 1.17 g (to three SF)

You could have done this as well:

58.443 g/mol times 0.0200 mol <--- this is based on knowing the answer from problem #7

Problem #9: What weight (in grams) of H2SO4 would be needed to make 750.0 mL of 2.00 M solution?

Solution:

(2.00 mol L¯1) (0.7500 L) = x / 98.0768 g mol¯1

x = (2.00 mol L¯1) (0.7500 L) (98.0768 g mol¯1)

x = 147.1152 g

to three sig figs, 147 g

Problem #10: What volume (in mL) of 18.0 M H2SO4 is needed to contain 2.45 g H2SO4?

Solution:

(18.0 mol L¯1) (x) = 2.45 g / 98.0768 g mol¯1

(18.0 mol L¯1) (x) = 0.0249804235 mol

x = 0.0013878 L

The above is the answer in liters. Multiplying the answer by 1000 provides the required mL value:

0.0013878 L times (1000 mL / L) = 1.39 mL (given to three sig figs)