(a) How many milliliters of a stock solution of \(6.0 \mathrm{MHNO}_{3}\) would you have to use to prepare \(110 \mathrm{~mL}\) of \(0.500 \mathrm{MHNO}_{3} ?\) (b) If you dilute \(10.0 \mathrm{~mL}\) of the stock solution to a final volume of \(0.250 \mathrm{~L},\) what will be the concentration of the diluted solution?

: We can use the dilution formula to solve this problem: M1V1 = M2V2 In this case, M1 = 6.0 M (stock solution concentration), M2 = 0.500 M (diluted solution concentration), and V2 = 110 mL (diluted solution volume). We want to find V1 (volume of stock solution required). Rearranging the formula for V1: V1 = M2V2 / M1 Substituting the known values: V1 = (0.500 M × 110 mL) / 6.0 M Calculating the volume: V1 = 9.167 mL Therefore, 9.167 mL of 6.0 MHNO3 stock solution would be required to prepare 110 mL of 0.500 MHNO3 solution.

: Again, we'll use the dilution formula for this part: M1V1 = M2V2 In this case, M1 = 6.0 M (stock solution concentration), V1 = 10 mL (stock solution volume used), V2 = 0.250 L (final volume of the diluted solution). We want to find M2 (concentration of the diluted solution). Rearranging the formula for M2: M2 = M1V1 / V2 Note that we need to make sure that our volume units are consistent. With V2 given in L, we will convert V1 to L. V1 = 10 mL = 0.010 L Now, we can substitute the values into the formula: M2 = (6.0 M × 0.010 L) / 0.250 L Calculating the concentration: M2 = 0.24 M Therefore, the concentration of the diluted solution after using 10.0 mL of the stock solution and a final volume of 0.250 L is 0.24 M.