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A student needs to dilute a 0.28 M Pb(NO3), solution to make 111.0 mL of 0.11 M Pb(NO3)2. Set up the calculation by placing t

I am wondering if this is right? Thanks

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Answer #1

The answer is, 43.60 ml of 0.28 M Pb(NO3)2 is needed.

We can use the formula V1N1=V2N2

It is given that N1 = 0.28 M

V2 = 111 mL

N2 =0.11 M

Hence V1 = V2N2 / N1

V1 = (111 mL × 0.11 M) / 0.28 M

= 43.60 mL

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