You make 1.000 L
of an aqueous solution that contains 35.0 g
of glucose (C6H12O6)
. How many liters of water would you have to add to this solution to reduce the molarity you calculated in Part A by a factor of two?
To find the molarity of the solution, we first need to calculate the number of moles of glucose present:
Molar mass of glucose (C6H12O6) = 180.16 g/mol
Number of moles of glucose = 35.0 g / 180.16 g/mol = 0.194 moles
The initial molarity of the solution is then given by:
Molarity = moles of solute / liters of solution
Molarity = 0.194 moles / 1.000 L = 0.194 M
To reduce the molarity by a factor of two, we need to find a dilution factor that would result in a final molarity of 0.097 M. Let's call this dilution factor "x".
Final molarity = 0.194 M / x = 0.097 M
x = 0.194 M / 0.097 M
x = 2
This means you need to double the volume of the solution. Since we started with 1.000 L, we would need to add an additional 1.000 L of water to reduce the molarity by a factor of two.