I fyou had only 12g of KOH remaining in a bottle, how many ml of 12.8%(w/v) solution could you prepare? (in mL)

How many ml of 0.29 M solution could you prepare using 12 g of KOH?

12.75 g less than the theoretically predicted mass of 25.69 g, what is the percent yield of this reaction?

To calculate the volume of a solution that can be prepared using a given amount of solute, you'll need to use the formula:

Volume (mL) = Mass (g) / Concentration (g/mL)

First, let's calculate the volume of 12.8% (w/v) KOH solution that can be prepared using 12g of KOH.

Step 1: Determine the concentration in g/mL

Since the concentration is given as 12.8% (w/v), it means that there are 12.8g KOH in 100 mL of solution.

Therefore, the concentration is 12.8g / 100 mL = 0.128 g/mL.

Step 2: Calculate the volume of the solution

To calculate the volume of the solution that can be prepared using 12g of KOH, divide the mass of KOH (12g) by the concentration (0.128 g/mL):

Volume (mL) = 12g / 0.128 g/mL = 93.75 mL

Therefore, with 12g of KOH, you can prepare approximately 93.75 mL of 12.8% (w/v) KOH solution.

Now, let's calculate the volume of 0.29 M KOH solution that can be prepared using 12g of KOH.

Step 1: Determine the molar mass of KOH

The molar mass of KOH can be calculated using the atomic masses of its constituent elements:

Molar mass (KOH) = Atomic mass (K) + Atomic mass (O) + Atomic mass (H)
= 39.10 g/mol + 16.00 g/mol + 1.01 g/mol
= 56.11 g/mol

Step 2: Determine the moles of KOH

To calculate the moles of KOH, divide the mass of KOH (12g) by its molar mass (56.11 g/mol):

Moles (KOH) = 12g / 56.11 g/mol = 0.214 moles

Step 3: Calculate the volume of the solution

The formula for molarity is:

Molarity (M) = Moles / Volume (L)

Rearranging the formula, we have:

Volume (L) = Moles / Molarity

Since we want the volume in mL, we can multiply the result by 1000:

Volume (mL) = (Moles / Molarity) * 1000

Substituting the values, we get:

Volume (mL) = (0.214 moles / 0.29 mol/L) * 1000 = 737.93 mL

Therefore, with 12g of KOH, you can prepare approximately 737.93 mL of 0.29 M KOH solution.

12.8% KOH is 12 g KOH/100 g solution; therefore, we could prepare

100g x 12.0/12.8 = 93.75 g soln.
Of the 93.75 g soln, 12.0 g would be KOH; therefore 93.75-12.0 = 81.75 g water + 12.0 g KOH.

12.0 g KOH is how many moles?
12/molarmass = 12/56.1 = 0.214 moles.
M = moles/L. Substitute to obtain
0.29M = 0.214/L and
L = 0.214/0.29 = 0.738 L which you can convert to mL.