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How many grams of KOH are present in 125.0 ml of a 5.75 M solution

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Answer:

there are 40.28 grams of KOH in 125.0 ml of a 5.75 M solution.

Step-by-step explanation:

To calculate the number of grams of KOH (potassium hydroxide) present in a solution, you can use the formula:

Grams of solute = (Molarity) x (Volume in liters) x (Molar mass)

First, convert the volume from milliliters to liters by dividing by 1000:

Volume in liters = 125.0 ml / 1000 = 0.125 liters

The molar mass of KOH is calculated by adding the molar masses of each element:

Molar mass of K (potassium) = 39.10 g/mol

Molar mass of O (oxygen) = 16.00 g/mol

Molar mass of H (hydrogen) = 1.01 g/mol

Molar mass of KOH = 39.10 + 16.00 + 1.01 = 56.11 g/mol

Now, you can use the formula to calculate the grams of KOH:

Grams of KOH = (5.75 M) x (0.125 L) x (56.11 g/mol)

Grams of KOH = 40.28 grams

So, there are 40.28 grams of KOH in 125.0 ml of a 5.75 M solution.

i hope this helps you

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