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Topic: Molarity of KCl Solution  (Read 22745 times)

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Offline Sis290025

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Molarity of KCl Solution
« on: January 18, 2009, 11:07:15 AM »
For the following, I would appreciate it if someone would check to see if my calculations and significant figures are correct.


1) For the calculation of the grams of KCl transferred to the beaker, my value is 0.3704 g.
 
a) Convert to grams of KCl per L of solution.
 
If the KCl that I transferred is 0.3704 g and the solution's volume is 0.250 L (250 mL), then
 
(0.3704 g)/(0.250 L - uncertainty in number of decimal places for this from volumetric flask) = 1.48 g of KCl per L (Did I use the correct number of significant figures?)
 
b) Calculate the molarity of KCl solution.
 
Given MW 74.56 g/mol.
 
(1.48 g KCl)*(1 mol KCl / 74.56 g/mol) = 0.0199 mol / L = 0.0199 M of KCl solution (Is this the correct number of significant figures?)
 
c) Find the concentration of solution in ppm of K.
 
Given KCl (aq) --> K+ (aq) + Cl- (aq)
 
Molarity calculated is 0.0199 mol/L for KCl.
 
0.0199 mol KCl * (1 mol K+ / 1 mol KCl) * (39.10 g K+ / 1 mol K+) = 0.778 g K+ / L
 
In ppm 0.778 g K+ * (1000 mg / 1 g) = 778 mg / L. (Correct significant figures?)
 


2) After finding the difference between my final and initial buret readings, the volume of the above KCl solution delivered was
 
47.85 mL - 33.67 mL = 14.18 mL.

14.18 mL was then diluted with water until it reached the mark on a 50 mL volumetric flask.

Calculate the molar concentration of the diluted KCl solution.
 
Using M_conc.*V_conc. = M_dil.*V_dil,
 
[(0.0199 M)*(14.18 mL)] / 50.00 mL (uncertainty in number of decimal places for volumetric flask) = 0.00564 M of diluted KCl solution (Correct number of significant figures?)

Thank you.

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