What volume (mL)of a 5.45 M lead nitrate solution must be diluted to 820.7 mL to make a 1.41 M solution of lead nitrate?
A)212
B)6310
C)3170
D)0.00936
E)0.00471

Question

Quizplus · Accepted Answer

We can use the formula:

M1V1 = M2V2

where:
M1 = initial molarity = 5.45 M
V1 = volume of initial solution to be diluted = unknown
M2 = final molarity = 1.41 M
V2 = final volume = 820.7 mL = 0.8207 L

Solving for V1:
V1 = (M2V2)/M1
V1 = (1.41 M)(0.8207 L)/(5.45 M)
V1 = 0.212 L = 212 mL

Therefore, 212 mL of the 5.45 M lead nitrate solution must be diluted to 820.7 mL to make a 1.41 M solution of lead nitrate.

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