A 5.95*g5.95⋅g mass of potassium bromide were dissolved in 400*cm^3400⋅cm3 water ... what was the concentration of the solution in mol*L^-1mol⋅L−1?

1 Answer
anor277
Jul 14, 2017

Approx. 0.1*mol*L^-10.1⋅mol⋅L−1......................

Explanation:

We use the quotient....."Molarity"-="Moles of solute"/"Volume of solution"Molarity≡Moles of soluteVolume of solution, and we know that 1*cm^3-=1*mL-=10^-3*L1⋅cm3≡1⋅mL≡10−3⋅L......

And thus "Molarity"=((5.95*g)/(119.0*g*mol^-1))/(400*cm^3xx10^-3*L*cm^-3)Molarity=5.95⋅g119.0⋅g⋅mol−1400⋅cm3×10−3⋅L⋅cm−3

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

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