4.5: Measuring Concentrations of Solutions

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Molarity

Exercise \(\PageIndex{1}\)

What is the molarity of an NaI solution that contains 4.5 g of NaI in 21.0 mL of solution?

1.4 M
0.030 M
0.0047 M
0.00014 M
0.21 M

Answer

a. 1.4 M

\(n_{NaI}=4.5 g NaI (\frac{1 mol}{149.89424})=0.03 mol NaI\)

\(M = \frac{mol}{L}=\frac{0.03}{21*10^{-3}L}=1.429 M\)

Exercise \(\PageIndex{2}\)

What volume of 0.742 M Na₂CO₃ solution contains 44.9 g of Na₂CO₃?

0.314 L
6.41 × 10³ L
0.571 L
3.53 × 10³ L
1.75 L

Answer: c. 0.571 L

Exercise \(\PageIndex{3}\)

How many moles of sulfate ions are there in a 0.650-L solution of 0.312 M Al₂(SO₄)₃?

0.203 mol
0.608 mol
6.25 mol
0.0676 mol
1.44 mol

Answer: b. 0.608 mol

Exercise \(\PageIndex{4}\)

What mass of Na₂CO₃ is present in 0.700 L of a 0.396 M Na₂CO₃ solution?

29.4 g
74.2 g
42.0 g
187 g
60.0 g

Answer: a. 29.4 g

Dilutions

Exercise \(\PageIndex{5}\)

In order to dilute 77.1 mL of 0.778 M HCl to 0.100 M, the volume of water that must be added is

67.2 mL
9.91 mL
6 * 10² mL
1.01 * 10^-3 mL
5.23 * 10² mL

Answer

e. 5.23 * 10² mL

\(M_{1}*V_{1}=M_{2}*V_{2}\)

\(V_{2}=\frac{\left (M_{1}*V_{1} \right )}{M_{2}}=\frac{\left ( 0.778 M * \left ( 77.1 * 10^{-3}L \right )\right )}{0.100 M}= .5998 L\)

Exercise \(\PageIndex{6}\)

A dilute solution is prepared by transferring 35.00 mL of a 0.6363 M stock solution to a 900.0 mL volumetric flask and diluting to mark. What is the molarity of this dilute solution?

0.02474 M
0.04949 M
0.1636 M
0.006186 M
0.3182 M

Answer: a. 0.02474 M

Exercise \(\PageIndex{7}\)

What volume of 1.27 M HCl is required to prepare 197.4 mL of 0.456 M HCl?

5.5 * 102 mL
3.41 * 102 mL
70.9 mL
0.0141 mL
1.14 * 102 mL

Answer: c. 70.9 mL