the molarity of the solution containing 2.8% solution of KOH is ?
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hello
M(KOH)=56 g/ml
w=
100% = 2.8%
c=n/V
n=m/M
ρ=1 g/mL.
Suppose, that m solution =100 g then
mKOH =w∙msolution
/100% = 2.8*100/100% = 2.8 g
nKOH=2.8/56 = 0.05 mol
c=0.05 mol/0.1 L = 0.5 mol/L = M/5
hope this helps....
M(KOH)=56 g/ml
w=
100% = 2.8%
c=n/V
n=m/M
ρ=1 g/mL.
Suppose, that m solution =100 g then
mKOH =w∙msolution
/100% = 2.8*100/100% = 2.8 g
nKOH=2.8/56 = 0.05 mol
c=0.05 mol/0.1 L = 0.5 mol/L = M/5
hope this helps....
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