mole fraction of solute in aqueous solution of 30% naoh
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Answered by
452
NaOH solution is prepared in water.
Molar mass of NaOH = 40 g
Mass of NaOH = 30 g
Molar mass of water = 18 g
Given mass of water = 70 g
Number of moles of NaOH = 3040 = 0.75 moles
Number of moles of H2O = 7018 = 3.88 moles
Mole fraction of NaOH = 0.75/0.75+3.88
=0.161
Molar mass of NaOH = 40 g
Mass of NaOH = 30 g
Molar mass of water = 18 g
Given mass of water = 70 g
Number of moles of NaOH = 3040 = 0.75 moles
Number of moles of H2O = 7018 = 3.88 moles
Mole fraction of NaOH = 0.75/0.75+3.88
=0.161
anitarawat220:
At last don't understood why 0.75/0.75+3.88
Please reply to my question
Oh I understood bcz we find mole fraction of naoh so at numerator we write moles of naoh and in denominator no.of moles of h2o+naoh
Answered by
211
30% NaOH means 100g of solution contains 30g of NaOH
Mass of water in solution = 100 - 30 = 70g
Moles of
NaOH = 30 / 40 = 0.75
H2O = 70 / 18 = 3.8
Mole fraction of NaOH = [(0.75) / (0.75 + 3.8)]
= 0.165
Mass of water in solution = 100 - 30 = 70g
Moles of
NaOH = 30 / 40 = 0.75
H2O = 70 / 18 = 3.8
Mole fraction of NaOH = [(0.75) / (0.75 + 3.8)]
= 0.165
thanks for your answer
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