# Calculate the concentration (in M) of Cl- ions in solution C

• AMan24
In summary, when mixing solutions A and B, the concentration of Cl- ions in solution C is 1.234 M, and the concentration of Na+ ions is 1.80 M.
AMan24

## Homework Statement

[/B]
PROBLEM #1 You have 3.00 L of a 3.39 M solution of NaCl(aq) called solution A. You also have 2.00 L of a 2.00 Msolution of AgNO3(aq) called solution B. You mix these solutions together, making solution C.

Calculate the concentration (in M) of Cl- ions in solution C.PROBLEM #2 You have 3.00 L of a 3.00 M solution of NaCl(aq) called solution A. You also have 2.00 L of a 2.00 Msolution of AgNO3(aq) called solution B. You mix these solutions together, making solution C.

Calculate the concentration (in M) of Na+ ions in solution C.

nope

## The Attempt at a Solution

For problem #2, i figured it out easily. I just did:
3.00M x 3.00L = 9moles.
9moles/(2.00L + 3.00L) = 1.80M

For problem #1, i found the answer but i don't understand why. Heres what its supposed to be like:

3.39M x 3.00L = 10.17moles Nacl.
2.00M x 2.00L = 4moles AgNO3 <------DON'T UNDERSTAND WHY IM FINDING BOTH FOR THIS ONE AND NOT THE OTHER

now you do 10.17 moles NaCl - 4moles AgNO3 = 6.17Moles, (no idea why I am subtracting for this and not the other).

Now divide 6.17 by the total like the other problem. 6.17/(2.00L + 3.00L) = 1.234M...

- i don't think its because one is a cation and one is an anion
- my guess is that its because AgCl is a solid and NaNO3 is aqueous

Last edited:
AMan24 said:
- my guess is that its because AgCl is a solid and NaNO3 is aqueous
Yes. Write down the balanced equation for what happens when you mix NaCl and AgNO3, with proper labels [(aq), etc.], and see what is in solution.

AMan24
AMan24 said:
For problem #2, i figured it out easily. I just did:
3.00M x 3.00L = 9moles.
9moles/(2.00L + 3.00L) = 1.80M

For the record: that's not a correct answer.

## 1. How do you calculate the concentration of Cl- ions in solution C?

To calculate the concentration of Cl- ions in solution C, you need to know the molar mass of Cl- (35.45 g/mol), the volume of solution C in liters, and the number of moles of Cl- present in solution C. You can then use the formula M = mol/L to calculate the concentration in units of mol/L or Molarity.

## 2. What is the molar mass of Cl-?

The molar mass of Cl- is 35.45 g/mol. This includes the atomic weight of chlorine (35.45 g/mol) and the charge of -1, since Cl- has one more electron than the neutral chlorine atom.

## 3. How do you determine the volume of solution C?

The volume of solution C can be determined by measuring the amount of solution using a graduated cylinder or other volumetric measuring tool. Make sure to take note of the units used (e.g. liters, milliliters) and convert if necessary.

## 4. What is the difference between molarity and molality?

Molarity (M) is a measure of concentration in units of moles per liter of solution, while molality (m) is a measure of concentration in units of moles per kilogram of solvent. Molarity is temperature-dependent, while molality is not affected by changes in temperature.

## 5. What are the units for concentration in this calculation?

The units for concentration are typically moles per liter (mol/L) or Molarity. However, you can also express concentration in other units, such as parts per million (ppm) or percentage (%).

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