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

Answer 1

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?

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Answer 2

Axel Alvarado

The concentration of the solution of potassium bromide in mol*L^-1 is approximately 0.0374 mol/L.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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

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