# Homework Help: Moles of silver perchlorate

1. Feb 8, 2006

### AngelShare

I was given several equations to use for this assignment so, before assuming that I got this right, I wanted to check my answers. I don't think I have them done correctly...

What is the molarity of a solution of silver perchlorate, AgClO4, if 3.2 moles of silver perchlorate are dissolved in 1.5 liters of water?

molarity = moles/ volume
molarity = 3.2/1.5
molarity = 2.13

How many moles of potassium iodide, KI, must be used to make 1.500 L of a 0.2 M solution?

moles = 1.500/0.2
moles = 7.5

What is the molarity of the solution produced when 151 g of sodium chloride, NaCl, is dissolved in enough water to prepare .375 L of solution?

molarity = grams/molar mass
molarity = 151/.375
molarity = 402.6

How many grams of potassium chloride, KCl, are needed to prepare 0.630 L of a 2.50 M solution of potassium chloride?

(2.50)(.630) = x/74.6
1.575= 74.6x
.021 = x
.021 grams

2. Feb 8, 2006

### ksinclair13

**Disclaimer**
I've learned from the past that I need to warn you ahead of time that I may be incorrect. I hope I don't accidentally mislead you.

What is the molarity of a solution of silver perchlorate, AgClO4, if 3.2 moles of silver perchlorate are dissolved in 1.5 liters of water?

Your work is correct, but make sure you check your significant figures.

How many moles of potassium iodide, KI, must be used to make 1.500 L of a 0.2 M solution?

M = mol / L ; mol = L * M

What is the molarity of the solution produced when 151 g of sodium chloride, NaCl, is dissolved in enough water to prepare .375 L of solution?

M = mol / L ; M = (g / molecular mass) / L

In other words, convert grams to moles first, and then plug that into the original equation.

How many grams of potassium chloride, KCl, are needed to prepare 0.630 L of a 2.50 M solution of potassium chloride?

You started it off right, but you finished it incorrectly. Think about it - you have moles, and you need to convert to grams. Using dimensional analysis, you would have to do:

mol ( g / mol) so that moles cancel leaving you with grams. So, what do you think you should do?

3. Feb 8, 2006

### Gokul43201

Staff Emeritus

Last edited: Feb 8, 2006
4. Feb 9, 2006

### GCT

kinsclair13 gave good explanations, for the third question you'll need to convert to moles, since molarity is moles/liter. The last one seems correct though, you'll need to find the moles, than simply convert from moles to grams using factor labeling.

5. Feb 9, 2006

### Gokul43201

Staff Emeritus
Actually, the only problem with the last one is an error in algebra. Double-check the math.

6. Feb 9, 2006

### AngelShare

Alright, let's see if I got this now... ^_^

1. 2.1...molarity? Is that how you put it?
2. .3 moles
3. 6.88 molarity
4. 117.4 grams

Last edited: Feb 9, 2006
7. Feb 9, 2006

### AngelShare

Yeah, that was pretty stupid.:rofl: I did it correctly on my paper the second time around and then, when I compared it to my first answer I thought, "How on Earth did I screw that up?":yuck: