Question

The student then takes a 1.00M stock solution of table sugar (sucrose C12H22O11) and mixes 0.305L of stock solution with additional distilled water to create a dilute solution with a total volume of 1.25L.
Explain how the student can determine the molarity of the resulting solution. Show a valid calculation for the final molarity. PLS HELP

Answer 1

Answer: The molarity of final solution is 0.244 M.

Explanation:

Given: $M_{1}$ = 1.00 M,       $V_{1}$ = 0.305 L

$M_{2}$ = ?,                $V_{2}$ = 1.25 L

Formula used to calculate the molarity f resulting solution is as follows.

$M_{1}V_{1} = M_{2}V_{2}$

Substitute the values into above formula as follows.

$M_{1}V_{1} = M_{2}V_{2}\\1.00 M \times 0.305 L = M_{2} \times 1.25 L\\M_{2} = \frac{1.00 M \times 0.305 L}{1.25 L}\\= 0.244 M$

Thus, we can conclude that the molarity of final solution is 0.244 M.