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Go to Molarity Problems #11-25

Go to Molarity Problems #26-35

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 H_{2}SO_{4} 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 Na_{2}CO_{3} 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 Na_{2}CO_{3} are there in 10.0 L of 2.00 M solution?

**Solution:**

M = moles of solute / liters of solution2.00 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 Na_{2}CO_{3} are in 10.0 mL of a 2.0 M solution?

**Solution:**

M = moles of solute / liters of solution2.0 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 Lx = 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 upx = 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 H_{2}SO_{4} 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 H_{2}SO_{4} is needed to contain 2.45 g H_{2}SO_{4}?

**Solution:**

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

^{1}) (x) = 0.0249804235 molx = 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)

Go to Molarity Problems #11-25