Calculate the molarity of 3 drops (0.15mL) of a 0.5M Na3PO4 solution after adding 15 drops (.75mL) of water.
(Presume drops are the same size, one drop = 0.05mL)((Presume volumes are additive))
So far I've figured .15g Na3PO4/163.94g/mol Na3PO4 = .0009moles
.75g H20/18.015 = 0.0416 mol H20
What are the next steps to solve this problem? Thank you!
you are diluting the sodium phosphate by a factor of .90/.15=6
that means, the new molarity is .5/6
Thanks Bob!
Curious where .90 comes from? .0009molX1000mL?
.5/6= .083M
To calculate the molarity, you need to determine the final volume of the solution after adding water.
Given that one drop is 0.05 mL, you can calculate the total volume of the solution by adding the volume of the Na3PO4 solution (3 drops x 0.05 mL/drop = 0.15 mL) and the volume of water added (15 drops x 0.05 mL/drop = 0.75 mL).
Total volume = 0.15 mL (Na3PO4 solution) + 0.75 mL (water) = 0.9 mL.
However, it is important to note that the volumes should be converted to liters to be consistent with the molarity units:
Total volume = 0.9 mL = 0.9 mL x (1 L/1000 mL) = 0.0009 L.
Now, you have determined the total volume of the solution. To calculate the molarity, you divide the moles of solute by the volume of the solution in liters.
Molarity (M) = moles of solute / volume of solution (in liters).
You have already calculated the number of moles of Na3PO4 as 0.0009 moles. Therefore, to find the molarity:
Molarity = 0.0009 moles / 0.0009 L = 1 M.
So, the molarity of the solution after adding the water is 1 M.