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Chapter 11: Problem 14

What volume of a \(0.580-M\) solution of \(\mathrm{CaCl}_{2}\) contains \(1.28 \mathrm{~g}\) solute?

Short Answer

Expert verified
The volume of a \(0.580-M\) solution of \(CaCl_{2}\) that contains \(1.28 g\) solute is approximately \(2.02 mL\).

Step by step solution

01

Calculate moles of solute

First, we need to convert the mass of the solute (1.28 g) to moles. For this, we have to find the molar mass of CaCl2. The molar mass of Ca (calcium) is 40.08 g/mol, and that of Cl (chlorine) is 35.45 g/mol. Molar mass of CaCl2 = 40.08 + (2 * 35.45) = 40.08 + 70.9 = 110.98 g/mol Now divide the mass of solute by the calculated molar mass: moles of solute = mass of solute / molar mass of CaCl2 moles = 1.28 g / 110.98 g/mol
02

Calculate volume of the solution

Now, we can use the provided molarity of the solution to find the volume: Molarity = moles of solute / volume of solution 0.580 M = (1.28 g / 110.98 g/mol) / volume Volume = (1.28 g / 110.98 g/mol) / 0.580 M After calculating, we get: Volume ≈ 0.00202 L Now convert the volume to milliliters: Volume ≈ 0.00202 L × (1000 mL / 1 L) ≈ 2.02 mL
03

Report the result

So, the volume of a 0.580-M solution of CaCl2 that contains 1.28 g solute is approximately 2.02 mL.

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