What is the molarity of .060 mol of #NaHCO_3# in 1500 mL of solution?

The number of moles of solute that are present must be determined first, and then the volume of solution (which must be in liters) must be found. Since we already know how many moles are present, all we need to do is convert the milliliters of solution to liters of solution using the conversion factor 1000 mL = 1 L. This gives you 1.5 L when you divide the 1500 mL that you are given by 1000 mL/L.

After that, you would just divide the two as follows:

Molarity is calculated as solute moles per liter of solution.

1.5 L of solution / 0.060 mol NaHCO3 = 0.040 M.

I hope this clarification is clear.

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Molarity ((M)) is calculated by dividing the number of moles of solute by the volume of solution in liters.

Given:

• Moles of NaHCO3 ((n)) = 0.060 mol
• Volume of solution ((V)) = 1500 mL = 1.5 L

(M = \frac{n}{V} = \frac{0.060 \text{ mol}}{1.5 \text{ L}})

(M ≈ 0.040 \text{ mol/L})