#"Solute 1"#, #"S1"#, has a molecular mass of #40.0gmol^-1#...what is...?

Question

...what is #[S1]# if a #4g# mass of this material is dissolved in a #200mL# volume of solution?

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Answer 1 · 2017-08-29T17:33:50

By definition, #"concentration"-="amount of stuff"/"volume of solution"#

And typically, we use #"molarity"-="moles of solute"/"volume of solution"#

And thus #"concentration of S1"#

#=((4*g)/(40.0*g*mol^-1))/(200*mLxx10^-3*L*mL^-1)=0.50*mol*L^-1#

And note the dimensionality of this solution....

We cancel out common terms in the numerator and denominator....

#=((4*cancelg)/(40.0*cancelg*mol^-1))/(200*cancel(mL)xx10^-3*L*cancel(mL^-1))=0.5*(1/(mol^-1))/L=0.50*mol*L^-1#

Why? Because #1/x^-1=1/(1/x)=x# as required................