What volume (ml)of a 3.45 M lead nitrate solution must be diluted to 450.0 ml to make a 0.990 M solution of lead nitrate?
A)129
B)109
C)101
D)56
E)45

Question

Quizplus · Accepted Answer

Let us use the formula for calculating the concentration of a diluted solution:

M1V1 = M2V2

where M1 and V1 are the concentration and volume of the initial solution, and M2 and V2 are the concentration and volume of the final solution.

We are given:

M1 = 3.45 M (concentration of the initial solution)
M2 = 0.990 M (concentration of the final solution)
V2 = 450.0 ml (volume of the final solution)

We can rearrange the formula to solve for V1, the volume of the initial solution that must be diluted:

V1 = (M2V2) / M1
V1 = (0.990 M x 450.0 ml) / 3.45 M
V1 = 128.57 ml

Rounding to the nearest whole number, we get:

V1 = 129 ml

Therefore, we need to dilute 129 ml of the 3.45 M lead nitrate solution to 450.0 ml to make a 0.990 M solution of lead nitrate.

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