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Question
If 4 g of \[\ce{NaOH}\] dissolves in 36 g of \[\ce{H2O}\], calculate the mole fraction of each component in the solution. Also, determine the molarity of solution (specific gravity of solution is 1g mL–1).
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Solution
Molar mass of \[\ce{NaOH}\] = 40 g/mole
Molar mass of water = 18 g/ mole
Mass of \[\ce{NaOH}\] = 4 g
Number of moles of NaOH = `(4 g)/(40 g)` = 0.1 mol
Number of moles of H2O = `(36 g)/(18 g)` = 2 mol
Mole fraction of water = `(26 g)/("Number of moles of NaOH" + "Number of moles of" H_2O)`
Mole fraction of NaOH = `2/(0.1 + 2) = 2/2.1` = 0.95
Number of moles of `2/0.1 = 0.1/2.1` = 0.047
Number of moles of \[\ce{NaOH}\]
Mass of solution = Mass of solute + Mass of solvent
Mass of \[\ce{NaOH}\] + Mass of water = 4 g + 36 g = 40 g
Specific gravity of solution = 1 g/ml
1 litre = 1000 ml volume of solution = 40 ml
∴ 40 ml = 0.04 litre
Molarity = `"Number of moles of solute"/"Volume in litre" = (0.1 "mol")/(0.04 "litre")` = 2.5 M
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