Question

calculate the molality and mole fraction of solute in aqueous solution containing 3.0 g of urea per 250 g of water

Answer 1

Answer: Molality(m) = 0.2m

              Mole fraction(X) = 0.0035

Step by Step Explanation:

Weight of Urea = 3g

Molar mass of Urea = 60 (NH2CONH2)

Molality(m) = (W×1000) ÷ (GMM×Mass of solvent)

                   =(3×1000) ÷ (60×250)

                   =0.2mol/g

Mole Fraction [X(urea)] = n(urea) ÷ [n(urea) + n(water)]

*[Here, n = no. of moles , Molar mass of water = 18]

                                        = (3/60) ÷ (3/60 + 250/18)

                                        = 0.0035 (unitless)

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