Question

A student calculated the molarity of a solution prepared by dissolving 0.730 mol of table sugar (sucrose, C12H22O11) in 1.8x10^3 mL of water as 4.06x10^-4 M C12H22O11. Explain the student's calculation error and explain how the student should solve for the correct value of molarity. Show a valid calculation for the molarity. II. The student then takes a 1.00 M stock solution of table sugar (sucrose, C12H22O11) and mixes 0.305 L of stock solution with additional distilled water to create a dilute solution with a total volume of 1.25 L. 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:

.879 M

Explanation:

Answer 2

Concept:

The number of moles of solute contained in one liter of solution is referred to as molarity, which is represented by the letter M.

Given:

Number of moles of table sugar = 0.730 mol

The volume of water = 1.8 * 10³ mL = 1.8 * 10³ * 10⁻³ L = 1.8 L

The concentration of stock solution of table sugar (M₂) = 1.00 M

The volume of the stock solution taken (V₂)= 0.305 L

The volume of the diluted solution (V₁) = 1.25 L

Find:

Explain the student's calculation error and how the student should solve for the correct molarity value.

Solution:

The molarity of the solution can be calculated as:

Molarity  = The number of moles of solute / The volume of solution (in liters)

By substituting the given values in the above expression we get the molarity of the solution as follows below.

                   Molarity = 0.730 mol / 1.8 L = 0.4055 mol/L= 0.41 M

The student errored by failing to convert a volume unit from milliliters to liters. After all, the definition of molarity is the number of solute molecules per liter of solution.

Now, applying the molarity equation,

                                      M₁V₁ = M₂V₂

             (diluted solution)           (stock solution)

By substituting the given values in the above expression we get the molarity of the diluted solution as follows below.

                              M₁ * 1.25 L = 1.00 M * 0.305 L

                                           M₁ = (1.00 M * 0.305 L)/ 1.25 L

                                           M₁ = 0.244 M

Thus, the final molarity of the diluted solution is 0.244 M.

Hence, the molarity of a solution prepared by dissolving 0.730 mol of table sugar (sucrose, C₁₂H₂₂O₁₁) in 1.8x10 mL of water as 4.06x10 M  C₁₂H₂₂O₁₁ is 0.41 M. The final molarity of the diluted solution is 0.244 M.

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