If an $11.9 \cdot g$ $11.9g$ mass of potassium chloride is dissolved in $256 \cdot m L$ $256mL$ of water, what is the concentration of the solution?

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Answer 1 · 2017-10-27T19:05:24

$0.624 M$ $"0.624 M"$

$molarity = \frac{moles of solute}{liters of solution}$ $"molarity" = ("moles of solute") / ("liters of solution")$

Its units are $mol/L$ $"mol/L"$ , or simply $M$ $"M"$ .

Convert $11.9 g KCl$ $"11.9 g KCl"$ to moles:

$11.9 cancel("g KCl") xx ("1 mol KCl") / (74.55 cancel("g KCl")) = "0.160 mol KCl"$

Convert milliliters to liters:

$"256 mL = 0.256 L"$

$"molarity" = "0.160 mol" / ("0.256 L") = "0.624 M"$

Answer 2 · 2017-10-27T19:17:48

$[KCl(aq)]=0.624*mol*L^-1$

$"Molarity"="Moles of solute"/"Volume of solution"$

And thus we can fill in this quotient....

$"Molarity"=((11.9*g)/(74.55*g*mol^-1))/(256*mL*xx10^-3*L*mL^-1)=0.624*mol*L^-1.$

Of course, we know in solution there are equimolar $K^+$ and $Cl^-$ ..

If an 11.9*g11.9⋅g11.9*g mass of potassium chloride is dissolved in 256*mL256⋅mL256*mL of water, what is the concentration of the solution?