Question

What is the molarity of a solution prepared by dissolving 12.0 g of sodium hydrogen carbonate, NaHCO3, in

water to make 250.0 mL of solution?

Answer 1

- 0.57138096 M

- 0.571mol

- 0.571 M NaHCO3

- 0.571 mol/NaHCO3

(these are just different ways to identify the units but they all work and mean the same thing)

Step-by-step explanation:

- molarity is measured by the ratio of the moles of a solute per liters of a solution

- the formula for molarity is:
`M = (mol (s) solute)/(L solution)`__________________________________________________________

- first, you need to convert 250mL to L so,
`250/1000=0.25`

- when it comes to converting g to mol you can do it in a separate equation but i prefer to do it all in one equation

__________________________________________________________

- so the equation for this specific problem would be set up like this:
`(12.0g)/(.25L) x (1mol)/(84.007 g) = x`

• 12.0g of NaHCO3
• .25 L of solution
• 1 mol: 84.007g (the molar mass of NAHCO3)

__________________________________________________________

- now to actually solve it:

`(12.0g)/(.25L) x (1mol)/(84.007 g) = (12.0)/(21.00175) = 0.57138096(or)/0.571mol/0.571 M NaHCO3/(0.571 mol/NaHCO3)`

• you just multiply across so
  `12x1=12` and
  `.25x84.007 = 21.00175` then divide so
  `12.0/21.00175 = 0.57138096`

hope this helps :)