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Pre Laboratory Exercise (see instructions p. 4): Include a calculation of the correct mass of H2salen needed (Mw = 268.3 g/mo
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Answer #1

Moles of copper ll chloride in 2 ml and 0.03M solution= molarity × volume in litre

= .03 mole/L × 2 × 10^-3 L

=6 ×10^-5 mole CuCl2

Because Copper ll Chloride and H2salen are in 1:1 ratio so moles of H2salen = 6×10^-5 mole.

As mole = mass / Molar mass

So mass of given compound H2salen =

= Moles × Molar mass

= 6 × 10^-5 mole × 268.3 gMole^-1

=16098 × 10^-5

=0.16097 g answer

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