What mass of glucose c6h12o6, must dissolve in 400 g of ethanol ch3ch2oh to make a 1.6 m solution?

  • Now the #"moles of solute"# are a constant. The volume of solution MAY change substantially with increasing or decreasing temperature. In some calculations #"molality"# is used in preference, which is defined by the quotient....

    #"molality"="moles of solute"/"kilograms of solvent"#

    ....this is temperature independent, and at lower concentrations, #"molarity"-="molality"#.

  • To get the molarity, you divide the moles of solute by the litres of solution.

    #"Molarity" = "moles of solute"/"litres of solution"#

    For example, a 0.25 mol/L NaOH solution contains 0.25 mol of sodium hydroxide in every litre of solution.

    To calculate the molarity of a solution, you need to know the number of moles of solute and the total volume of the solution.

    To calculate molarity:

    1. Calculate the number of moles of solute present.
    2. Calculate the number of litres of solution present.
    3. Divide the number of moles of solute by the number of litres of solution.

    EXAMPLE:

    What is the molarity of a solution prepared by dissolving 15.0 g of NaOH in enough water to make a total of 225 mL of solution?

    Solution:

    1 mol of NaOH has a mass of 40.00 g, so

    #"Moles of NaOH" = 15.0 cancel("g NaOH") × "1 mol NaOH"/(40.00 cancel("g NaOH")) = "0.375 mol NaOH"#

    #"Litres of solution" = 225 cancel("mL soln") × "1 L soln"/(1000 cancel("mL soln")) = "0.225 L soln"#

    #"Molarity" = "moles of solute"/"litres of solution" = "0.375 mol"/"0.225 L" = "1.67 mol/L"#

    Some students prefer to use a "molarity triangle".

    What mass of glucose c6h12o6, must dissolve in 400 g of ethanol ch3ch2oh to make a 1.6 m solution?

    It summarizes the molarity formulas as

    #"Moles" = "molarity × litres"#

    #"Molarity" = "moles"/"litres"#

    #"Litres" = "moles"/"molarity"#