how many ML of each of the following solutions will provide 25.0g of KOH? A)2.50 M of KOH b)0.750 M of KOH c)5.60 M of KOH NEED HELP setting up problem!!

I would do this.

L x M = moles and moles x molar mass = grams; therefore, L x M x molar mass = grams. Substitute and solve for L in each case, then convert L to mL.

thanks

To determine the volume of each solution needed to obtain 25.0g of KOH, we can use the formula:

Volume (in mL) = (Mass of KOH) / (Concentration in Molarity)

Let's go through each part:

a) 2.50 M KOH solution:
Volume (in mL) = 25.0g / 2.50 mol/L
= (25.0g / 2.50) / (1 L / 1000 mL)
= (25.0g / 2.50) * (1000 / 1) mL
= 250 mL

So, you will need 250 mL of the 2.50 M KOH solution.

b) 0.750 M KOH solution:
Volume (in mL) = 25.0g / 0.750 mol/L
= (25.0g / 0.750) / (1 L / 1000 mL)
= (25.0g / 0.750) * (1000 / 1) mL
= 1111.111 mL (rounded to three decimal places)

So, you will need approximately 1111.111 mL of the 0.750 M KOH solution.

c) 5.60 M KOH solution:
Volume (in mL) = 25.0g / 5.60 mol/L
= (25.0g / 5.60) / (1 L / 1000 mL)
= (25.0g / 5.60) * (1000 / 1) mL
= 4464.286 mL (rounded to three decimal places)

So, you will need approximately 4464.286 mL of the 5.60 M KOH solution.