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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?

Respuesta :

Answer:

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

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