Asked by Anonymous
                If you recovered 4.65 g of Cu from a 42.0 mL sample of material, what would the molarity be?
4.65 g Cu/ 63.5 g/mol = 0.0732 mol
0.0732 mol/ 42 mL x 1000 mL/ 1 L =1.74 M ~ 2 M Cu??? Is this correct sig fig because you go off of the 1 sig in 1,000 mL??? So you get 2 with rounding 1.74 with following sig fig rules..
            
        4.65 g Cu/ 63.5 g/mol = 0.0732 mol
0.0732 mol/ 42 mL x 1000 mL/ 1 L =1.74 M ~ 2 M Cu??? Is this correct sig fig because you go off of the 1 sig in 1,000 mL??? So you get 2 with rounding 1.74 with following sig fig rules..
Answers
                    Answered by
            DrBob222
            
    I would say M = mols/L
mols - 0.0732
L - 0.0420
M - 1.74M and use all three digits because, by definition, 1000 mL = 1 L.
    
mols - 0.0732
L - 0.0420
M - 1.74M and use all three digits because, by definition, 1000 mL = 1 L.
                    Answered by
            Anonymous
            
    oh, but with 1 L it is still only one sig fig? right?
    
                    Answered by
            DrBob222
            
     Yes, 1 L has only 1 s.f.; however, 42.0 mL = 0.0420 L and there are 3 s.f. in 4.65g and in 0.0420. The "1 L" is never used since the definition is M = mols/L. The "L" in the formula happens to be 0.0420.
    
                    Answered by
            Anonymous
            
    Okay that makes sense but if you had 40 mL which is just .04 then it would just be 1 sig fig.
    
                    Answered by
            Anonymous
            
    .04 L to be exact
    
                    Answered by
            DrBob222
            
    As I undersand it, yes. 0.04 has 1 s.f. 
    
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