Question

A solution is prepared by dissolving 29.2 g of glucose (C6H12O6) in 355 g of water. The final volume of the solution is 378 mL . For this solution, calculate each of the following.

molarity

molality

percent by mass

mole fraction

mole percent

Answer 1

1. Calculation of molarity

      weight of glucose = 29.2 grams

      volume of solution = 378 ml = 378 *10^-3 Litre

     molar mass of glucose = 180 grams/mole

      Number of moles of glucose = 29.2/180 = 0.1622 moles

Molarity is given by ,

Molarity(M) = moles of solute / Volume of solution ( in Litres)

                = 0.1622 / ( 378 *10^-3)

                = 0.4916 M

Molarity of the solution is 0.4916 M

2. calculation of molality

        Weight of solvent = 355 grams =0.355 Kg

        Number of moles of solute ( glucose) = 0.1622 moles

Molality is given by,

molality (m) = moles of solute/ Kg of solvent

   = 0.1622 / 0.355

= 0.4569 m

molality of given solution is 0.4569 m

3.) percent by mass :

     To calculate percent by mass, we will have to calculate weight fraction and than multiply weight fraction by 100 to obtain percent by mass

Mass of glucose = 29.2 grams

mass of water = 355 grams

Total mass = mass of glucose + mass of water = 29.2 +355 = 384.2 grams

mass fraction of glucose = mass of glucose / total mass = 29.2 / 384.2 = 0.076

    mass percent of water = mass of water / total mass = 355/384.2 = 0.9239

    percent by mass of glucose = mass fraction of glucose * 100 =0.076 *100 = 7.6 %

   percent by mass of water = mass fraction of water *100 = 0.9239*100 = 92.39 %

4.Mole fraction

   First we will have to calculate moles of glucose and water

     mass of glucose = 29.2 grams

     mass of water = 355 grams

    molar mass of glucose = 180 grams/mole

    molar mass of water = 18 grams/mole

    moles of glucose = mass of glucose / molar mass of glucose = 29.2/180 = 0.1622 moles

    moles of water = mass of water / molar mass of water = 355/18 = 19.722 moles

      Total moles = moles of glucose + moles of water = 0.1622+19.722 = 19.8844 moles

      Moles fraction of glucose = moles of glucose / total moles = 0.1622/19.8844 = 0.008157

     mole fraction of water = moles of water / total moles = 19.722/19.8844 = 0.99183

4. Mole percent

    Mole percent of glucose = mole fraction of glucose * 100 = 0.008157 *100 =0.8157 %

   mole percent of water = mole fraction of water *100 =0.99183*100 = 99.183 %

     Final answer

Molarity of the solution is 0.4916 M

molality of given solution is 0.4569 m

percent by mass : percent by mass of glucose = mass fraction of glucose * 100 =0.076 *100 = 7.6 %

   percent by mass of water = mass fraction of water *100 = 0.9239*100 = 92.39 %        

Mole fraction :   Moles fraction of glucose = moles of glucose / total moles = 0.1622/19.8844 = 0.008157

   mole fraction of water = moles of water / total moles = 19.722/19.8844 = 0.99183

Mole percent : Mole percent of glucose = mole fraction of glucose * 100 = 0.008157 *100 =0.8157 %

   mole percent of water = mole fraction of water *100 =0.99183*100 = 99.183 %