A solution is made up by weighing out 20g of sodium ethanoate, NaO2CCH3 dissolving in water and making up to?

Question

A solution is made up by weighing out 20g of sodium ethanoate, NaO2CCH3 dissolving in water and making up to a volume of 250mL.

(a) How many moles of sodium ethanoate were used?

Answer 1

Atomic weights: C = 12 H = 1 O = 16 Na = 23 NaC2H3O2 = 83

Let sodium ethanoate be called sodium acetate and symbolized SA

20gSA x 1molSA/83gSA = 0.24molSA to two significant figures

20g is the given. 83 is the molecular weight. The grams SA cancel, leaving moles SA.

Answer 2

Molarity is without worry purely the moles of the solute (something you're dissolving) divided by way of way of the liters of the finished determination. to locate the moles of the sodium acetate, you basically divide the grams by way of the molar mass. 4.00g of sodium acetate divided by way of applying 80 two.0 g of sodium acetate in one mole can provide approximately 0.0488 moles of sodium acetate. Now, commerce the mL to L (once you communicate approximately that that's the ideal way it is interior the equation) and 750 mL = 0.seventy 5 L. at last, adjust to the 1st equation i gave, which became into as quickly as moles divided by way of liters and you get.... 0.0488 moles / 0.seventy 5 L = 0.0.5-molar determination