Question 7.C-L.6: Describe how you would prepare the following, using pure sol...
Describe how you would prepare the following, using pure solute and water:
a. 500 mL of 1.00 M MgCl_2 solution
b. 100 mL of 12.0% (w/v) MgCl_2 solution
c. 1.00 L of 20.0% (v/v) ethylene glycol solution
a. M = \frac{mol of solute}{liters of solution}
mol of solute = (M) (liters of solution)
= (1.00 M) (0.500 L)
= 0.500 mol of solute
The solute is MgCl_2, which has a formula weight of 95.2 u. The formula weight provides the factor in the following calculation:
0.500 \cancel{mol} \cancel{MgCl_2} × \frac{95.2 g MgCl_2}{1 \cancel{mol} \cancel{MgCl_2}} = 47.6 g MgCl_2The solution is prepared by putting 47.6 g of MgCl_2 into a 500-mL flask and adding pure water to the mark, making certain all the solid solute dissolves and the resulting solu-tion is well mixed.
b. \%(w/v) = \frac{g of solute}{mL of solution} × 100
g of solute = \frac{\%(w/v) (mL of solution)}{100} \\ \\ = \frac{(12.0\%) (100 mL)}{100} = 12.0 gThe solution is prepared by putting 12.0 g of MgCl_2 into a 100-mL flask and adding pure water to the mark, making certain all the solid solute dissolves and the resulting solution is well mixed.
c. \%(v/v) = \frac{solute volume}{solution volume} × 100
solute volume = \frac{\%(v/v) (solution volume)}{100} \\ \\ = \frac{(20.0\%) (1.00 L)}{100} = 0.200 L \\ \\ = 200 mLThe solution is prepared by putting 200 mL of ethylene glycol into a 1.00-L flask and adding pure water to the mark, making certain the two liquids are completely mixed to form the final solution.