A student makes a solution by dissolving 55.8 grams of potassium hydroxide in 875.0 grams of water. The resulting solution has a density of 1.07 grams per milliliter.

1. What is the volume of this solution?

2. Calculate the % concentration of this solution.

3. Calculate the molar concentration of this solution.

4. How many grams of potassium hydroxide are there in 150.0ml of this solution?

5. A 200.0ml sample of this solution is diluted to 500.0ml with water. What is its molar concentration?

6. What volume of this solution contains 12.75 grams of potassium hydroxide?

7. What mass of the original solution contains 25.0 grams of potassium hydroxide?

8. What volume of 0.750M H2SO4 solution can be neutralized by 100.0 grams of this solution?

9. A student needs 0.600L a 0.100M solution of potassium hydroxidde. Describe how this can be done using the original solution and a 1000ml graduated cylinder.

1. To find the volume of the solution, divide the mass of the solution by its density.

Volume = Mass / Density
Volume = 55.8g + 875.0g = 930.8g
Density = 1.07g/mL
Volume = 930.8g / 1.07g/mL = 869.7 mL

2. To calculate the % concentration of the solution, divide the mass of potassium hydroxide by the mass of the solution, and multiply by 100.
% Concentration = (Mass of Potassium Hydroxide / Mass of Solution) x 100
% Concentration = (55.8g / 930.8g) x 100 = 5.99%

3. To calculate the molar concentration of the solution, we need to know the molar mass of potassium hydroxide (KOH). The molar mass of KOH is:
K: 39.10 g/mol
O: 16.00 g/mol
H: 1.01 g/mol
Molar mass of KOH = 39.10 + 16.00 + 1.01 = 56.11 g/mol

Now, we can calculate the molar concentration using the following formula:
Molar Concentration = (Mass of Potassium Hydroxide / Molar Mass of Potassium Hydroxide) / Volume
Molar Concentration = (55.8g / 56.11 g/mol) / 0.8697 L = 0.9916 mol/L

4. To find the grams of potassium hydroxide in 150.0 mL of the solution, we can use the % concentration.
Grams of Potassium Hydroxide = (150.0 mL / 100 mL) x (55.8g / 100 mL) = 83.7g

5. When the 200.0 mL sample is diluted to 500.0 mL, the number of moles of the solute (potassium hydroxide) remains the same. Therefore, the molar concentration remains the same.
Molar Concentration = 0.9916 mol/L

6. To find the volume of the solution that contains 12.75 grams of potassium hydroxide, we can rearrange the formula for molar concentration and solve for volume.
Volume = (Mass of Potassium Hydroxide / Molar Mass of Potassium Hydroxide) / Molar Concentration
Volume = (12.75g / 56.11 g/mol) / 0.9916 mol/L = 0.228 L

7. To find the mass of the original solution that contains 25.0 grams of potassium hydroxide, we can again use the % concentration.
Mass of Solution = (25.0 g / 0.0599) = 417 g

8. To determine the volume of 0.750M H2SO4 solution that can be neutralized by 100.0 grams of this solution, we can use stoichiometry.
Balanced equation: KOH + H2SO4 → K2SO4 + H2O
From the balanced equation, we know that 1 mole of KOH reacts with 1 mole of H2SO4.

First, we need to calculate the number of moles of KOH in 100.0 grams:
Moles of KOH = (100.0 g / 56.11 g/mol) = 1.78 mol

Since the stoichiometry is 1:1, we need 1.78 moles of H2SO4 to neutralize the KOH.
Molarity = Moles / Volume
Volume = Moles / Molarity
Volume = 1.78 mol / 0.750 mol/L = 2.37 L

9. To make a 0.100M solution of potassium hydroxide with a 1000mL graduated cylinder, we will need to calculate the volume of the original solution needed.
Molarity = Moles / Volume
Volume = Moles / Molarity
Volume = (0.100 mol/L) x (0.600 L) = 0.060 mol

Now, we need to find the corresponding mass of potassium hydroxide using the molar mass.
Mass of Potassium Hydroxide = (0.060 mol) x (56.11 g/mol) = 3.37 g

Since the original solution has a % concentration of 5.99%, we can calculate the volume of the original solution needed:
Volume of Original Solution = (Mass of Potassium Hydroxide / % Concentration) x 100
Volume of Original Solution = (3.37 g / 5.99%) x 100 = 56.2 mL

Therefore, to obtain a 0.100M solution of potassium hydroxide, you would need to measure 56.2 mL of the original solution using a 1000mL graduated cylinder and then dilute it to 600 mL with water.

1. To find the volume of the solution, we can use the formula:

Volume = Mass / Density

Given that the mass of the solution is the sum of the mass of potassium hydroxide and water (55.8 g + 875.0 g) and the density is 1.07 g/mL, we can substitute these values into the formula:

Volume = (55.8 g + 875.0 g) / 1.07 g/mL

Simplifying the expression:

Volume = 930.8 g / 1.07 g/mL

Volume ≈ 869.35 mL

Therefore, the volume of this solution is approximately 869.35 mL.

2. To calculate the % concentration of the solution, we need to find the mass of potassium hydroxide and divide it by the total mass of the solution, then multiply by 100.

% Concentration = (Mass of KOH / Total mass of solution) x 100

Given that the mass of potassium hydroxide is 55.8 g and the total mass of the solution (found in question 1) is 930.8 g, we can substitute these values into the formula:

% Concentration = (55.8 g / 930.8 g) x 100

% Concentration ≈ 6.00 %

Therefore, the % concentration of this solution is approximately 6.00%.

3. To calculate the molar concentration of the solution, we need to find the number of moles of potassium hydroxide and divide it by the volume of the solution in liters.

Molar concentration (M) = Moles of solute / Volume of solution (in liters)

First, we need to find the number of moles of potassium hydroxide. We can use the formula:

Moles = Mass / Molar Mass

Given that the molar mass of potassium hydroxide is 56.11 g/mol and the mass of potassium hydroxide is 55.8 g, we can substitute these values into the formula:

Moles = 55.8 g / 56.11 g/mol

Moles ≈ 0.994 mol

Next, we convert the volume of the solution from milliliters to liters:

Volume = 869.35 mL * (1 L / 1000 mL)

Volume ≈ 0.86935 L

Finally, we substitute the moles of potassium hydroxide and the volume of the solution into the formula:

Molar concentration (M) = 0.994 mol / 0.86935 L

Molar concentration ≈ 1.14 M

Therefore, the molar concentration of this solution is approximately 1.14 M.

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