A solution is prepared by adding 0.300 g of solid NaCl to 50.0 mL of 0.100 M CaCl2. What is the molarity of chloride ion in the final solution? Assume that the volume of the final solution is 50.0 mL.

The molarity of the chloride ion is 0.3026 M ≈ 0.303 M

Explanation:

Step 1: Data given

Mass of solid NaCl = 0.300 grams

Volume of CaCl2 = 50.0 mL

Molarity of CaCl2 solution = 0.100 M

Volume of the final solution = 50.0 mL

Step 2: The balanced equation

NaCl  → Na+ + Cl-

CaCl2  → Ca^2+ + 2Cl-

CaCl2 → NaCl2 + CaCl

Step 3: Calculate the number of moles Cl2 in CaCl2

In 1 mol CaCl2 we have 1 mol Ca^2+ and 2 mol Cl-

Moles CaCl2 = molarity ¨volume

Moles CaCl2 = 0.100 M * 0.050 L

Moles CaCl2 = 0.005 moles

In 1 mol CaCl2 we have 1 mol Ca^2+ and 2 mol Cl-

For 0.005 moles CaCl2 we have 2*0.005 = 0.010 moles Cl-

Step 4: Calculate moles Cl in NaCl

Moles NaCl = mass NaCl / molar mass NaCl

Moles NaCl = 0.300 grams / 58.44 g/mol

Moles NaCl = 0.00513 moles

In 1 mol NaCl we have 1 mol Na+ and 1 mol Cl-

In 0.00513moles NaCl we have 0.00513 moles Cl-

Step 5: Calculate the total number of moles Cl-

Total moles Cl- = 0.010 moles + 0.00513 moles

Total moles Cl- = 0.01513 moles Cl-

Step 6: Calculate the molarity of Cl-

Molarity = moles / volume

Molarity Cl- = 0.01513 moles / 0.050 L

Molarity Cl- = 0.3026 M

The molarity of the chloride ion is 0.3026 M ≈ 0.303 M

a pan containing water is placed in a positio n where it is exposed to a 10 mile per hour wind. the pan contains a uniform depth of 1.27 cm of water, is quite wide, and has a length of 2.44 m in the wind direction. the water in the pan is at a constant temperature of 15.5 °c.