A 0.380 m aqueous solution of NaCl is prepared at 20.0∘C. Assume that the density of...

A 0.380 m aqueous solution of NaCl is prepared at 20.0∘C. Assume that the density of the solution at 20.0∘C is 1.082 g/mL.

Calculate the molarity of the salt solution.

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Expert Solution

Answer :

Molality m = # of moles of solute / 1 kg of solvent.

Mass of solute dissolved in 1 Kg solvent = molality x Molar mass of solute.

Molar mass of NaCl = 58.5 g and molality m = 0.380 m

Hence mass of NaCl dissolved in 1kg of solvent water = 0.380 x 58.5 = 22.23 g.

Hence total mass of 0.380 m NaCl solution = 22.23 g solute + 1000 g water = 1022.23 g = 1.02223 kg.

Density of NaCl solution = 1.082 g/mL

Using mass of NaCl solution and density let us calculate volume of the NaCl solution.

Volume = Mass /Density = 1022.23 / 1.082 = 944.76 mL = 0.94476 L

It means 22.23 g i.e. 0.380 moles of NaCl are disolved and diluted to 944.76 mL i.e. 0.94476 L.

Hence molarity ofNaCl is ,

Molarity = # of moles of solute / volume of solution in L

Molarity of NaCl = 0.380 / 0.94476 = 0.402 M

Molarity of 0.380 m NaCl solution at 20 ^oC is 0.402 M.

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queen_honey_blossom answered 1 year ago

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