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Chemistry Forums for Students => Analytical Chemistry Forum => Topic started by: Arathor on December 03, 2014, 08:20:34 PM

Title: Standard Dilution Calculation
Post by: Arathor on December 03, 2014, 08:20:34 PM

I always mange to mess up theses questions.
The answer is 647.2 ppm

5.672 g CaCl₂ transferred into a 100.0 mL VF (Topped). 17.86 mL of this delivered into a 1.000 L volumetric flask (Topped)

Calculate the concentration of chloride ion in the 1 L volumetric flask (PPM)

(5.972g/110.98g/mol)*.1L=0.05381 M
(0.05381 M)(0.01786L)/(1L)= 0.000961 M

This is probably where I go wrong
9.61E-4 CaCl₂ (mM of Cl /mM ofCaCl₂)
=3.07E-4moles of Cl- *(35.45g/mol Cl)
=0.0109g of Cl- (1000mg)
=10.883mg / 1L = 10.883mg

Title: Re: Standard Dilution Calculation
Post by: Hunter2 on December 04, 2014, 01:09:16 AM

In the first flask you have 5.672 g /100 ml = 56.72 g/l = 56.72 mg/ml.
You take 17.86 ml will give you 1013.02 mg This is in 1 l means 1.013 g/l CaCl₂

In mol it gives 1,013 g/l / 111 g/mol = 0,0091 mol/l CaCl₂

Cl^- is then the double what means 0,0182 mol.

Calculated the mass of Chloride will give you 0.642 g/l = 642 mg/l = 642 ppm

Your calculation :

Quote

(5.972g/110.98g/mol)*.1L=0.05381 M

is wrong

It has to be (5.972g/110.98g/mol)/0.1L=0.5381 M

It will guide to the same result. Your last calculations I don't understand.