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An aqueous sodium acetate, NaC2H3O2 , solution is made by dissolving 0.395 mol NaC2H3O2 in 0.505 kg of water. Calculate the molality of the solution.

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  • BatteringRam


    Answer: The molality of NaC_2H_3O_2 solution is 0.782 m

    Explanation:

    Molality is defined as the amount of solute expressed in the number of moles present per kilogram of solvent. The units of molarity are mol/kg. The formula used to calculate molality:

    \text{Molality of solution}=\frac{\text{Moles of solute}}{\text{Mass of solvent (in kg)}} .....(1)

    Given values:

    Moles of NaC_2H_3O_2 = 0.395 mol

    Mass of solvent (water) = 0.505 kg

    Putting values in equation 1, we get:

    \text{Molality of }NaC_2H_3O_2=\frac{0.395mol}{0.505kg}\\\\\text{Molality of }NaC_2H_3O_2=0.782m

    Hence, the molality of NaC_2H_3O_2 solution is 0.782 m