# Stumped on a Molarity problem

• antonisz

## Homework Statement

Calculate the molarity of a solution made by adding 35.5 mL of concentrated ammonia (28.0 % by mass, density 0.880 g/mL) to some water in a volumetric flask, then adding water to the mark to make exactly 250 mL of solution. (It is important to add concentrated acid or base to water, rather than the other way, to minimize splashing and maximize safety.)
u

## Homework Equations

Molarity = mol / liter

Mass % = (grams of solute / grams of solution) * 100

Density = mass / volume

## The Attempt at a Solution

a) I used the ml of NH3 to find the number of grams of NH3 - 35.5 ml * 0.88 g/ml = 31.24 g

b) I then used that amount to find the number of mole - (31.24 g / 17.03) = 1.8344 mol of NH3

This is where I am stuck. I used the % mass formula to find that the total grams of solution is 111.57 (0.28 = 31.24 / x and cross multiply), but I have no idea where to go from here.

I know that if I compute the molarity using (1.8344 mol / .250 L) I'll get the moles of NH3 at concentration 100%, would I need to multiply the number of moles by 0.28 to get the 28% concentration?

*Edit* Yes, it turns out that that would make sense, too much chemistry for the last 2 hours! I haven't taken Gen Chem in a couple years and getting thrown into Analytical, I should have reviewed some of these concepts. [/B]

antonisz said:

## Homework Statement

Calculate the molarity of a solution made by adding 35.5 mL of concentrated ammonia (28.0 % by mass, density 0.880 g/mL) to some water in a volumetric flask, then adding water to the mark to make exactly 250 mL of solution. (It is important to add concentrated acid or base to water, rather than the other way, to minimize splashing and maximize safety.)
u

## Homework Equations

Molarity = mol / liter

Mass % = (grams of solute / grams of solution) * 100

Density = mass / volume

## The Attempt at a Solution

a) I used the ml of NH3 to find the number of grams of NH3 - 35.5 ml * 0.88 g/ml = 31.24 g

b) I then used that amount to find the number of mole - (31.24 g / 17.03) = 1.8344 mol of NH3

This is where I am stuck. I used the % mass formula to find that the total grams of solution is 111.57 (0.28 = 31.24 / x and cross multiply), but I have no idea where to go from here.

I know that if I compute the molarity using (1.8344 mol / .250 L) I'll get the moles of NH3 at concentration 100%, would I need to multiply the number of moles by 0.28 to get the 28% concentration?

*Edit* Yes, it turns out that that would make sense, too much chemistry for the last 2 hours! I haven't taken Gen Chem in a couple years and getting thrown into Analytical, I should have reviewed some of these concepts. [/B]

In step (a), you have found the mass of ammonia solution . The mass of ammonia molecules in this solution is 28% of this. Remember that ammonia is a gas, the stuff you are working with is ammonia gas dissolved in water. Taking 28% of 31.24 g will tell you how many grams of ammonia are actually present in the solution. Continue on as you have done to calculate the final concentration.

antonisz said:

## Homework Statement

Calculate the molarity of a solution made by adding 35.5 mL of concentrated ammonia (28.0 % by mass, density 0.880 g/mL) to some water in a volumetric flask, then adding water to the mark to make exactly 250 mL of solution. (It is important to add concentrated acid or base to water, rather than the other way, to minimize splashing and maximize safety.)
u

## Homework Equations

Molarity = mol / liter

Mass % = (grams of solute / grams of solution) * 100

Density = mass / volume

## The Attempt at a Solution

a) I used the ml of NH3 to find the number of grams of NH3 - 35.5 ml * 0.88 g/ml = 31.24 g

b) I then used that amount to find the number of mole - (31.24 g / 17.03) = 1.8344 mol of NH3

This is where I am stuck. I used the % mass formula to find that the total grams of solution is 111.57 (0.28 = 31.24 / x and cross multiply), but I have no idea where to go from here.

I know that if I compute the molarity using (1.8344 mol / .250 L) I'll get the moles of NH3 at concentration 100%, would I need to multiply the number of moles by 0.28 to get the 28% concentration?

*Edit* Yes, it turns out that that would make sense, too much chemistry for the last 2 hours! I haven't taken Gen Chem in a couple years and getting thrown into Analytical, I should have reviewed some of these concepts. [/B]
I think you are making a mistake while calculating the mass of ammonia. The solution contains 28% by mass of ammonia. 35.5 x .88 will give you the mass of the solution, not NH3. Multiply that further by 0.28 to get the grams of ammonia. Then retry solving the problem.

siddharth23 said:
I think you are making a mistake while calculating the mass of ammonia. The solution contains 28% by mass of ammonia. 35.5 x .88 will give you the mass of the solution, not NH3. Multiply that further by 0.28 to get the grams of ammonia. Then retry solving the problem.
Yes, thank you. I caught my mistake. Just some careless thinking on my part.

Thank you for the response.

antonisz said:
Yes, thank you. I caught my mistake. Just some careless thinking on my part.

Thank you for the response.