Question 4.8: Calculating the Mass of Solute in a Solution of Known Molari......
Calculating the Mass of Solute in a Solution of Known Molarity
What mass of K_{2}CrO_{4} is needed to prepare exactly 0.2500 L (250.0 mL) of a 0.250 M K_{2}CrO_{4} solution in water? (See Figure 4-5.)
Analyze
The conversion pathway is L soln → mol K_{2}CrO_{4} → g K_{2}CrO_{4}. The first conversion factor is the molarity of the solution, shown below in blue, and the second conversion factor is the molar mass of K_{2}CrO_{4}.
Solve
? g K_{2}CrO_{4} = 0.2500 L soln × \frac{0.250 mol K_{2}CrO_{4}}{1 L soln } × \frac{194.2 g K_{2}CrO_{4}}{1 mol K_{2}CrO_{4}}
= 12.1 g K_{2}CrO_{4}
Assess
The answer has the correct units. We can check whether the answer is reasonable by working the problem in reverse and using numbers that are rounded off slightly. Because the molar mass of K_{2}CrO_{4} is approximately 200 g/mol and the mass of the sample is approximately 12 g, the number of moles of K_{2}CrO_{4} in the sample is approximately 12/200 = 6/100 = 0.06 mol. The approximate molarity is 0.06/0.250 = 0.24 mol/L. This estimate is close to the true molarity, and so we are confident that the answer, 12.1 g K_{2}CrO_{4}, is correct.