We are being asked of the **volume **of 3.45 M lead(II) nitrate solution be diluted in order to make 450.0 mL of a 0.990 M solution of lead(II) nitrate.

When dealing with **dilution **we will use the following equation:

$\overline{){{\mathbf{M}}}_{{\mathbf{1}}}{{\mathbf{V}}}_{{\mathbf{1}}}{\mathbf{}}{\mathbf{=}}{\mathbf{}}{{\mathbf{M}}}_{{\mathbf{2}}}{{\mathbf{V}}}_{{\mathbf{2}}}}$

M_{1} = initial concentration

V_{1} = initial volume

M_{2} = final concentration

V_{2} = final volume

To what * volume (in mL)* must a 3.45 M lead(II) nitrate solution be diluted in order to make 450.0 mL of a 0.990 M solution of lead(II) nitrate?

A. 129

B. 109

C. 101

D. 56.0

E. 45.0

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