How many grams of NaC2H3O2 are needed to prepare 350 ml of a 2.75 M solution?

Answer 1

First find the number of moles of the sodium bicarbonate by dividing the mass you have(12.5g) by the molar mass of the compound(84.0g) to get .149 moles.

Then divide the number of moles (.149mol) by the number of liters of solute (.350L) to get the molarity which is .426 M or .426 mol/L to the correct number of significant figures

Molarity = Grams/(Molecular Weight X Volume)

Given:

grams = 12.5g ; molecular weight of NAHCO3 = 84.00661 g/mol ; volume =350ml or 0.35L

Solution

Molarity = 12.5 g / ( 84.00661 g/mol x 0.35 L)

= 12.5 g / ( 29.4023135 g/mol L )

= 0.425136614 mol L or approx 0.43 mol L

Answer 2

79.0 g NaC2H3O2 I think