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A solution is prepared by dissolving 28.4 g of glucose (C6H1206) in 355 g of water. The final volume of the solution is 378 ml. For this solution, calculate the concentration in each unit: 3. Molarity b. Molality c. Mole fraction of glucose d. Mole fraction of water

Respuesta :

Given, Mass of Glucose= 28.4g

           Mass of Water   = 355g or 0.355 Kg

           Volume of solution= 378ml or 0.378 litres

So        Moles of Glucose = 28.4/180.16 mol

                                         = 0.157 mol

           Moles of Water     = 355 / 18 mol = 19.7 mol

Hence         Molarity = Moles of Glucose / volume of the solution

                                   = 0.157 / 0.378 M = 0.417 M

                     Molality = Moles of Glucose /  Mass of Water(Kg)

                                    = 0.157 / 0.355 m = 0.44 m

                     Mole fraction of glucose =  

               Moles of Glucose / (  Moles of Glucose +  Moles of Water )

                                    = 0.157 / (0.157+ 19.7)

                                    = 0.0079

                     Mole fraction of water    =

              Moles of Water / (  Moles of Glucose +  Moles of Water )

                                    =  19.7 / (0.157+ 19.7)

                                     = 0.9920

What is solution?

The following are several characteristics of solutions:

  • It is a uniform blend.
  • With a diameter of less than 1 nm, its particles are excessively small to the eye, the particles are invisible.
  • Since particles don't scatter light when it passes through them, the path of the light cannot be seen.
  • The mixture cannot be separated from the solutes, and they do not settle. A stable solution exists.
  • Filtration cannot separate the components of a mixture.

Learn more about solution: https://brainly.com/question/20388765

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