A 5.00 mL solution of BaCl_2 dissolved in water weighed 5.227 g. After the water

Question

A 5.00 mL solution of BaCl_2 dissolved in water weighed 5.227 g. After the water was removed by heating 0.236 g of BaCl_2 remained. Using the formulas on page 38 in the lab manual calculate the following knowing that BaCl_2 is the component. What Is the density of the BaCl_2 solution in g/mL? Percent by mass of the BaCl_2 In the solution: Concentration of BaCl_2 in the solution in g/mL: How many mole BaCl_2 in the solution. The formula mass of BaCl_2 is 208.24 g/mol, (use mol as unit for mole)? What is the molarity of the BaCl_2 solution in mol/L?

Explanation / Answer

5.00 mL of the BaCl2 solution weighs 5.227 g. The density of the solution is given by

Density = (mass of solution)/(volume of solution) = (5.227 g)/(5.00 mL) = 1.0454 g/mL (ans).

5.227 g of the solution furnished 0.235 g of BaCl2 on heating. Therefore, the solution contained 0.235 g BaCl2 and (5.227 – 0.235) g = 4.992 g water.

Percent of BaCl2 in the solution = (mass of BaCl2)/(mass of solution)*100 = (0.235 g)/(5.227 g)*100 = 4.4958 4.496% (ans).

We have 0.235 g BaCl2 dissolved in 5.00 mL of water (assume volume of BaCl2 to be negligible). The concentration of BaCl2 in g/mL of the solution = (mass of BaCl2)/(volume of solution) = (0.235 g)/(5.00 mL) = 0.047 g/mL (ans).

We have 0.235 g BaCl2. Molar mass of BaCl2 is 208.24 g/mol.

Therefore, moles of BaCl2 = (mass of BaCl2 taken)/(molar mass of BaCl2) = (0.235 g)/(208.24 g/mole) = 0.001128 mole (ans).

Volume of the solution in L = (5.00 mL)*(1 L/1000 mL) = 0.005 L.

Molar concentration of BaCl2 = (moles of BaCl2)/(volume of solution in L) = (0.001128 mole)/(0.005 L) = 0.2256 mol/L = 0.2256 M (ans).

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