Ask Question
29 January, 00:25

Suppose you have an aqueous solution containing 158.2 g KOHper

liter, its density is 1.13 g/cm3. You want to prepare a0.250 molal

solution of KPH, starting with 100.0 mL of the originalsolution.

How much water or solid KOH should be added to the100.0 mL

portion?

+5
Answers (1)
  1. 29 January, 00:39
    0
    1132.8 ml of water

    Explanation:

    you have an aqueous solution contains 158.2 g KOH per liter

    so concentration = 158.2/56 = 2.825M

    Molarity = 2.825

    that means you have 2.825 moles of KOH in 1.00L solution

    Mass of Soluet (KOH) = 152.8g

    Volume of solution = 1.00L

    density of solution = 1.13g/cm3 = 1.13g/ml

    therefore mass of solution = VolumeX density = 1000mL X 1.13g/ml.=1130g

    Mass of solvent (water) = mass of solution - mass of solute (KOH) = 1130-152.8 = 997.2g

    Molality = moles of solute/mass of solvent (Kg)

    =2.825 / (997.2/1000) = 2.832molal

    to prepare a 0.250 molal solution of KOH, starting with 100.0ml ofthe orginal solution

    0.250*X = 2.832 * 100

    X = 1132.8 ml of water you have to add
Know the Answer?
Not Sure About the Answer?
Find an answer to your question 👍 “Suppose you have an aqueous solution containing 158.2 g KOHper liter, its density is 1.13 g/cm3. You want to prepare a0.250 molal solution ...” in 📗 Chemistry if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers