4.5: Measuring Concentrations of Solutions
- Page ID
- 166324
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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