Please help me to solve the first, the second, the third and the fourth question
ID: 474631 • Letter: P
Question
Please help me to solve the first, the second, the third and the fourth questions. Thank you so much. The following questions all refer to an aqueous 7.69 M potassium dichromate stock solution. 1. If 17.1 mL of the stock solution are diluted to 480. m what is the molarity (in mol/L) of the new solution? Submit Answer Tries 0/3 2. If 28.3 mL of the stock solution are diluted to obtain a 0.474 M potassium dichromate solution, what is the volume (in mL) of the new solution? Submit Answer Tries 0/3 3. Determine the volume (in mL) of the original stock solution that must be used to prepare 661 mL of a 0.482 M potassium dichromate solution. Submit Answer Tries 0/3 4. Determine the volume of water (in mL) that was added to 21.6 mL of the stock solution to prepare a 0.719 M potassium dichromate solution. Submit Answer Tries 0/3Explanation / Answer
M1V1 = M2V2
where M1 and M2 is the molarity of the original and diluted dichromate solution respectively. V1 and V2 is the volume of the original and diluted solution.
M2 = M1*V1/V2
= 7.69 M *17.1 mL/ 480 mL
= 0.274 M
Molarity of the diluted solution is 0.274 M
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M1V1 = M2V2
where M1 and M2 is the molarity of the original and diluted dichromate solution respectively. V1 and V2 is the volume of the original and diluted solution.
V2 = M1*V1/M2
= 7.69 M *28.3 mL /0.474 M = 459.13 mL
Volume of the new solution = 459.13 mL
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M1V1 = M2V2
where M1 and M2 is the molarity of the original and diluted dichromate solution respectively. V1 and V2 is the volume of the original and diluted solution.
V1 = m2*V2/M1
= 0.482 M *661 mL/7.69 M = 41.43 mL
41.43 mL original solution will be required.
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M1V1 = M2V2
where M1 and M2 is the molarity of the original and diluted dichromate solution respectively. V1 and V2 is the volume of the original and diluted solution.
V2 = M1*V1/M2
= 21.6 mL * 7.69 M/0.719 M = 213.02mL
volume of water that has to be added= 213.02-21.6 = 209.42 mL