How much dry solute would you take to prepare each of the following solutions fr

Question

How much dry solute would you take to prepare each of the following solutions from the dry solute and the solvent?

D.) How much solvent would you take to prepare the solution in part B?

E.) How much solvent would you take to prepare the solution in part C?

part A) i got 5.21g KCl

part C) i got 5.5g KCl

pleaseeeeeeeee help me with parts B D E !!!!!!!   :D

How much dry solute would you take to prepare each of the following solutions from the dry solute and the solvent? 126 mL of 0.555M KCl 124 g of 0.440m KCl 124g of 4.4% KCl solution by mass How much solvent would you take to prepare the solution in part B? How much solvent would you take to prepare the solution in part C? i got 5.21g KCl i got 5.5g KCl

Explanation / Answer

B) molality = n of solute/mass of solvent

==> 0.440 = (mass/(39+35.5))/(124-m)*10^-3

==> (124-m)*0.440*(39+35.5) = 1000m

==> 4064.72 -32.78*m = 1000m

==> 1032.78m = 4064.72

==> m =4064.72/1032.78 = 3.94 g

D) mass of solvent = 124-3.94 = 121.06 g

E)mass of solvent = 124-5.5 = 118.5 g

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