here is the question: describe how you would prepare thefollowing solution: 100g
ID: 691409 • Letter: H
Question
here is the question: describe how you would prepare thefollowing solution: 100g of .500m KCli got this part: .500mol KCl/1 kg H2O X 74.55 g KCl/1 mol KClX 1 kg H2O/1000g h20= .03728 g KCl/g H2O
after that i am not totally sure how to get to grams of H20for the solution.
i don't understand how we go from gH20= gKCl -gsolution to theratio .03728=x g KCl/ x g KCL -100g solution. what does that ratioreally mean? please help. here is the question: describe how you would prepare thefollowing solution: 100g of .500m KCl
i got this part: .500mol KCl/1 kg H2O X 74.55 g KCl/1 mol KClX 1 kg H2O/1000g h20= .03728 g KCl/g H2O
after that i am not totally sure how to get to grams of H20for the solution.
Explanation / Answer
We know thatmolality m = ( mass / Molar mass ) / Mass of the solvent inKgMolar mass of KCl = 39 + 35.5 = 74.5 g / mol
Given molality m = 0.5 m mass of the solute KCl m' = 100 g Plug the values we get mass of solvent M' =( mass / Molar mass ) / molality = ( 100 g / 74.5 g / mol ) / 0.5 m = 2.6845 Kg density of water d = 1 g / mL Density d = mass / Volume Therefore Volume of water V = mass / density = 2.6845 * 1000 g / 1 g / mL = 26845 mL = 2.6845 L So take 100 g of the of KCl and dissolved in 2.6845 L of water.
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