Converting moles of a substance to milliliters of the same substance. You must know the substance's atomic or molecular weight and its density to do this conversion. Here's the basic strategy:
moles | molecular weight |
grams | density |
milliliters |
For example, to find the number of milliliters that 10.0 moles of NaCl would occupy, you must find the molecular weight (58.45 g/mol) and look up the density (2.17 g/mL):
10.0 mol NaCl | ( | 58.45 g NaCl 1 mol NaCl | ) | ( | 1 mL NaCl 2.17 g NaCl | ) | = 269 mL NaCl |
Converting moles of solute to milliliters of solution. You'll need to know the concentration of the solute in the solution to do this conversion. For example, to find the number of milliliters of 0.123 M AgNO_{3} solution that contains 1.00 mol of AgNO_{3}, convert moles of AgNO_{3} to liters of solution using the molarity:
1.00 mol AgNO_{3} | ( | 1 L solution 0.123 mol AgNO_{3} | ) | ( | 1000 mL solution 1 L solution | ) | = 8.13×10^{3} mL |
Converting moles of gas to milliliters of gas. If you know the density of the gas, you can use the first conversion method to do this type of problem. But if you don't know the density, and the gas can be treated as ideal, you can calculate the molar volume of the gas from the ideal gas law and use it to convert moles to liters directly. Since PV=nRT, V/n = RT/P. For example, to find the number of milliliters that 1.00 mol of helium would occupy at 298 K and 1.00 atm, compute the molar volume
V n |
= | RT P |
= | ( | 0.0821 L atm mol^{-1} K^{-1}× 298 K 1.00 atm | ) | = 24.4658 L/mol |
1.00 mol gas | ( | 24.4658 L gas mol gas | ) | ( | 1000 mL gas 1 L gas | ) | = 2.45×10^{4} mL gas |
Copyright © 1997-2010 by Fred Senese
Comments & questions to fsenese@frostburg.edu
Last Revised 08/17/15.URL: http://antoine.frostburg.edu/chem/senese/101/moles/faq/print-moles-to-liters.shtml