1.) B.)To what volume must 200. mL of 4.50 M Al(NO3)3 be diluted to produce 3.00...

Question

Question

1.)

B.)To what volume must 200. mL of 4.50 M Al(NO3)3 be diluted to produce 3.00 M Al(NO3)3?

A.) If 155 mL of 1.70 M KCl are diluted to 650. mL, what is the molarity of the final solution?

Answer 1

Answer #1

use dilution formula
M1*V1 = M2*V2
1---> is for stock solution
2---> is for diluted solution

B)
M1*V1 = M2*V2
4.50 M* 200 mL = 3 M*V
M = 692 mL
Answer: 692 mL

A)
M1*V1 = M2*V2
1.70 M*155 mL = M*650 mL
M = 0.405 M
Answer: 0.405 M

Answer 2

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