The density of 3 M solution is 1.25 g mL-1. Calculate molality of the solution.
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The ques. is incomplete ... NaCl should be given...
Molarity = no. of moles of solute / volume of solution in L
Given, no. of moles of NaCl = 3
Density = Mass/ Volume
Mass of the solutuon = density x volume
= 1.25 x 1000 = 1250 g
Mass of solute = mass of 3 moles of NaCl = 3 x (23 +35.5) = 175.5 g
∴ mass of solvent = mass of solution - mass of solute
= 1250 - 175.5 = 1074.5 g =1.074 kg
∴ molality = no. of moles of solute/ mass of solvent = 3/1.074 = 2.79
∴ molarity = (molality x density x 1000 )- molar mass /1000
= (2.79 x 1.25 x 1000) - 58.5 /1000 = 3.4M
Molarity = no. of moles of solute / volume of solution in L
Given, no. of moles of NaCl = 3
Density = Mass/ Volume
Mass of the solutuon = density x volume
= 1.25 x 1000 = 1250 g
Mass of solute = mass of 3 moles of NaCl = 3 x (23 +35.5) = 175.5 g
∴ mass of solvent = mass of solution - mass of solute
= 1250 - 175.5 = 1074.5 g =1.074 kg
∴ molality = no. of moles of solute/ mass of solvent = 3/1.074 = 2.79
∴ molarity = (molality x density x 1000 )- molar mass /1000
= (2.79 x 1.25 x 1000) - 58.5 /1000 = 3.4M
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