A $5.95g$ mass of potassium bromide were dissolved in $400cm^3$ water ... what was the concentration of the solution in $mol*L^-1$ ?

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Answer 1 · 2017-07-14T07:40:07

Approx. $0.1*mol*L^-1$ ......................

We use the quotient..... $"Molarity"-="Moles of solute"/"Volume of solution"$ , and we know that $1*cm^3-=1*mL-=10^-3*L$ ......

And thus $"Molarity"=((5.95*g)/(119.0*g*mol^-1))/(400*cm^3xx10^-3*L*cm^-3)$

$=0.125*mol*L^-1$ with respect to $KBr$ . What are the concentrations with respect to $K^+$ and $Br^-$ ? Are the units I use correct?

A 5.95*g5.95*g mass of potassium bromide were dissolved in 400*cm^3400*cm^3 water ... what was the concentration of the solution in mol*L^-1mol*L^-1?