More dilute sodium hydroxide solutions are often prepared from a sastuated sodium hydroxide soltuion that's is 52.0% by mass and has a density of 1.48 g/mL. Calculate the volume, in liters, of this solution needed to prepare 35.0 L if 0.125 M sodium hydroxide?

52.0% w/w means 52 g NaOH/100 g solution.

Convert 52 g NaOH to moles. moles = grams/molar mass.
Use the density to convert 100 g soln to mL, then to L. Then molarity = moles/L soln. Finally, mL x M = mL x M.

To calculate the volume of the sodium hydroxide solution needed to prepare 35.0 L of a 0.125 M solution, we need to use the equation:

Molarity (M) = Moles of solute (mol) / Volume of solution (L)

First, we need to calculate the number of moles of sodium hydroxide (NaOH) required for the 0.125 M solution:

Molarity (M) = 0.125 M
Volume of solution (L) = 35.0 L

Rearranging the equation, we can calculate the moles of NaOH:

Moles of solute (NaOH) = Molarity (M) * Volume of solution (L)
Moles of solute (NaOH) = 0.125 M * 35.0 L
Moles of solute (NaOH) = 4.375 moles

Now, we need to determine the mass of sodium hydroxide required to obtain 4.375 moles. To do this, we will use the given information about the concentration and density of the saturated sodium hydroxide solution.

The solution is 52.0% by mass, meaning that 100 g of the solution contains 52 g of sodium hydroxide (NaOH).

Density of the solution = 1.48 g/mL

To calculate the mass of NaOH required, we need to find the volume of the saturated sodium hydroxide solution that contains 4.375 moles of NaOH:

Volume of solution (mL) = Moles of solute (NaOH) / Molarity (M)
Volume of solution (mL) = 4.375 mol / 52 g/mol (molar mass of NaOH)
Volume of solution (mL) ≈ 84.135 mL

Next, we need to convert the volume of the solution from milliliters (mL) to liters (L):

Volume of solution (L) = 84.135 mL / 1000 mL/L
Volume of solution (L) ≈ 0.084 L

Therefore, approximately 0.084 L of the saturated sodium hydroxide solution is required to prepare 35.0 L of a 0.125 M sodium hydroxide solution.