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Question
The density of 3 molal solution of \[\ce{NaOH}\] is 1.110 g mL–1. Calculate the molarity of the solution.
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Solution
3 molal solution of \[\ce{NaOH}\] means 3 moles of \[\ce{NaOH}\] is dissolved in 1000 g of water.
But 3 moles of \[\ce{NaOH}\] = 3 × 40 g = 120 g
120 g = 1120 g
Density of solution = 1.110 g mL–1
∴ Volume of solution = `"Mass"/"Density" = (1120 g)/(1.11 g
"mL"^-1` = 1009 mL = 1.009 L
Molarity of the solution = `"Moles of solute"/"Volume of solution in L" = (3 "mole")/(1.009 "L")` = 2.97 M
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