Question

study the following sketch of a molecular orbital

Suppose 0.407 g of sodium acetate is dissolved in 100. mL of a 44.0 mM aqueous solution of ammonium sulfate. 

Answer 1

moles of sodium acetate (CH?COONa) = 0.407/82 = 0.00496 mole

so, moles of sodium cation = 0.00496 mole

final volume of solution = 100 ml = 0.1 L

so final molarity of sodium cation = 0.00496/0.1 = 4.96 x 10^-2 M

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