Adding 100-g HOH = 100-ml HOH into 10.6-M NaOH => 10.6 mole/Liter + 0.100L HOH => 10.6 mole NaOH/1.10 Liters Solution => 9.64 Molar in NaOH.
H2SO4 + 2NaOH => Na2SO4 + 2HOH
1/2(9.64M) H2SO4 + 9.64M NaOH => 1/2(Na2SO4) + HOH
=> (4.82mol/L)H2SO4 + (9.64mol/L)NaOH
=> Neutral Solution.
Using moles H2SO4 = 4.82 mole, and
moles of NaOH = 9.64 ... The mole fraction can be computed as follows...
X(NaOH) = 9.64 mole / (4.82 + 9.64) total moles = 0.6667 mole fraction NaOH ...
X(Na2SO4) = 4.82 mole / (4.82 + 9.64) total moles = 0.3333 mole fraction Na2SO4
Sum mole fractions = 1 = 0.6667NaOH + 0.3333Na2SO4.
100g of water is added into 10.6M NaOH solution. The diluted NaOH then neutralized completely with H2SO4 . density of NaOH is 1.08g/ml . calculate the mole fraction of NaOH in diluted solution .
1 answer
1 answer