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calculate the molarity,molality and mole fractionof 60% naoh solution of density 1.2g/cm3

shifa shahina , 9 Years ago
Grade 11
anser 1 Answers
Bhavya

Last Activity: 9 Years ago

60% is percentage by weight since nothing is given which implies that 60g of NaOH is present in 100g of solution.
Molality = (moles of solute/mass of solvent)*1000
moles of NaOH =weight of solute/Molecular mass = 60/40 =1.5
mass of solvent = mass of solution-mass of solute = 100-60 =40 g
m = (1.5/40)*1000
= 37.5m
For molarity we have a short cut method :-
Molarity = (10*x*d)/M.M
x=% by weight
d= density
M.M = Molecular mass
M = (10*60*1.2)/40
= 18M
mole fraction of solute(NaOH) = moles of solute/(moles of solute + moles of solvent)
= {1.5/(1.5+40/18)}
= 0.4032

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