[alt-photo] Re: molarity problem

[Alberto Novo]

Joseph, 

> KBr is 119 gm/mol
> KI is 166 gm/mol
> CdBr2 is 272 gm/mol
> CdI2 is 366 gm/mol

CdBr2 is 272.22
CdI2 is 366.22 

> Therefore the substitution would be ~ 1 gm KBr / 1.14 gm CdBr2.   
> (Similarly, the substitution between KI and CdI2 would be ~1 gm KI /  1.10 
> gm CdI2.)
> If a formula called for 1.0 gm CdBr2, I would need to substitute ~  0.88 
> gm KBr.

1 g CdBr2 is equivalent to 1*2*KBr/CdBr2 (2 because there are two Br in 
CdBr2)= 0.874292851 KBr (round to the nearest decimal you can weight).
1 g Kbr is 1/0.874... = 1.143781513 CdBr2 

1 g CdI2 is, with the same ratios, 0.906558899 KI, etc. 

Alberto
www.grupponamias.com
www.alternativephotography.com/articles/art102.html