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Chapter 4: Problem 74

(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 \(\mathrm{M} \mathrm{HNO}_{3} ?\) (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?

Short Answer

Expert verified
(a) To prepare 110 mL of 0.500 M HNO3 solution, we need 9.17 mL of the 6.0 M HNO3 stock solution. (b) The concentration of the diluted solution when 10.0 mL of the stock solution is diluted to 0.250 L is 0.240 M.

Step by step solution

01

Part (a): Finding the volume of the stock solution

Use the dilution formula to find the volume of the stock solution required to prepare the 0.500 M HNO3 solution: M1V1 = M2V2 We know: M1 = 6.0 M (concentration of the stock solution) M2 = 0.500 M (concentration of the diluted solution) V2 = 110 mL (volume of the diluted solution) Rearrange the formula to find V1: V1 = (M2V2) / M1 Plug in the given values: V1 = (0.500 M × 110 mL) / 6.0 M Calculate the value of V1: V1 = 9.17 mL So, to prepare 110 mL of 0.500 M HNO3 solution, we need 9.17 mL of the 6.0 M HNO3 stock solution.
02

Part (b): Finding the concentration of the diluted solution

Use the dilution formula to find the concentration of the diluted solution when 10.0 mL of the stock solution is diluted to 0.250 L: M1V1 = M2V2 We know: M1 = 6.0 M (concentration of the stock solution) V1 = 10.0 mL (volume of the stock solution) V2 = 0.250 L = 250 mL (volume of the diluted solution) Rearrange the formula to find M2: M2 = (M1V1) / V2 Plug in the given values: M2 = (6.0 M × 10.0 mL) / 250 mL Calculate the value of M2: M2 = 0.240 M So, the concentration of the diluted solution when 10.0 mL of the stock solution is diluted to 0.250 L is 0.240 M.

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Most popular questions from this chapter

Formic acid, HCOOH, is a weak electrolyte. What solutes are present in an aqueous solution of this compound? Write the chemical equation for the ionization of HCOOH.

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