1. In 100 mL of a solution, there are 3.0 g of NaCl. Find the molarity.
3.0 g NaCl (mole/58.5 g) = 0.05128205128 mole NaCl
n = CV
0.05128205128 mole = C(0.100 L)
C = 0.05128205128 mole/(0.100 L) = 0.51 moles/L
2. In 2.0 L of a solution, there are 3.0 g of KF. Find the molarity.
3.0 g of KF(mole/58 g)= 0.05172413793 mole KF
n = CV
0.05172413793 mole = C(2.0 L)
C = 0.05172413793 mole/2.0 L
= 0.026 moles/L
3. How many grams of KBr are needed to prepare 2.5 L of a 0.25 mole/L solution?
n = CV
= 0.25 mole/L (2.5 L)
= 0.625 moles KBr
0.6258 moles KBr *(39.1+79.90)g/mole = 75 g (74.5)
4. How many grams of Ca(ClO)2 are needed to prepare 2.0 L of a 0.45 mole/L
solution?
n = CV
= 0.45 mole/L (2.0 L)
= 0.90 moles Ca(ClO)2
0.90 moles Ca(ClO)2 (142.98 g/mole) = 128.682 g
5. What is the volume of a solution containing 3.0 grams of HNO3 if the concentration is 0.10 mole/L.
n = CV
First need n:
3.0 g HNO3 (mole/(1+14+48)) = 3/63 = 0.04761904762 moles HNO3
n = CV
0.04761904762 moles HNO3 = (0.10 mole/L)V
V = 0.04761904762 moles HNO3/ 0.10 mole/L = 0.48 L
6. Explain how you would actually prepare 3.0 L of a 0.2 mole/L Na Br solution in the lab.
n = CV
= 0.2 mole/L Na Br(3.0 L)
=0.6 moles Na Br
0.6 moles Na Br (102.89g/mole)
= 61.7 g
Weigh 61.7 g
Dissolve in < 3.0 L in a beaker
Transfer to a 3.0 Lvolumetric flask
Add water to white line and mix
Don’t forget the rest of the homework