Asolution is prepared by adding 1.70 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 chloride ions in the solution is 0.78 M

Explanation:

To calculate the number of moles, we use the equation:

Given mass of NaCl = 1.70 g

Molar mass of NaCl = 58.5 g/mol

Putting values in above equation, we get:

1 mole of NaCl produces 1 mole of sodium ions and 1 mole of chloride ions.

Moles of chloride ions = 0.029 moles

To calculate the number of moles for given molarity, we use the equation:

    .....(1)

Molarity of calcium chloride solution = 0.100 M

Volume of solution = 50.0 mL

Putting values in equation 1, we get:

1 mole of calcium chloride produces 1 mole of calcium ions and 2 moles of chloride ions.

Moles of chloride ions = (2 × 0.005) = 0.01 moles

Now, calculating the molarity of chloride ions in final solution, we use equation 1:

Total moles of chloride ions in solution = (0.029 + 0.01) = 0.039 moles

Total volume of solution = 50.0 mL

Putting values in equation 1, we get:

Hence, the molarity of chloride ions in the solution is 0.78 M

b) mutualism

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