In: Chemistry
If mL of solution is mixed mL of another solution of same substance, the Molarity of resulting solution is
Given that
0.05 L of 1.0 M lemonade, 0.05 L of 2.5 M lemonade, and .05 L of 0.5 M lemonade solutions are mixed.
Concentration of resultant solution is
M = 1.333 M
Molarity of resultant solution is 1.33 M
This problem can also be solved in the following method.
Number of moles of Lemonade in 0.05 L of 1 M solution is
= 0.05 L * 1 M = 0.05 moles
Number of moles of Lemonade in 0.05 L of 2 M 2.5 solution is
= 0.05 L * 2.5 M = 0.125 moles
Number of moles of Lemonade in 0.05 L of 0 .5 M solution is
= 0.05 L * 0.5 M = 0.025 moles
Total number of moles of Lemonade in the resultant solution is
= 0.05 moles + 0.125 moles + 0.025 moles
= 0.2 moles
Total volume of the solution is
0.05 L + 0.05 L + 0.05 L = 0.15 L
M = Number of moles of solute/ Volume of solution in Litres
M = 0.2 moles /0.15 L
M = 1.33 moles /L
M = 1.33 M