Question

What is the molality of a solution that has been prepared by mixing 52.1 g of sodium chloride with 479 g of water?

Answer 1

The molality of solution prepared by adding 52.1 g of NaCl to 479 g of water is 1.87 M.

Calculation:

We know that

Molality = Number of moles of solute/ Weight of solvent in Kg

And,

Number of moles = given mass/molar mass

Given,

Mass of NaCl = 52.1 g

Weight of solvent (water) = 479 g

                                         = 0.479 Kg

To find,

Molality of solution =?

Molar mass of NaCl = 23+35.5

                                 = 58.5 g/mol

Number of moles of NaCl = 52.1/58.5

                                           = 0.89 moles

                                           = 0.9 moles

Molality = Number of moles of solute/ Weight of solvent in Kg

             = 0.9/0.479

             = 1.87 M

Hence, the molality of the given solution is 1.87 M.

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