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What volume of 0.171 M NaOH, in milliliters, contains 29.0 g of NaOH?

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Answer: 4240.08

Explanation: 29 grams of NaOH is equivalent to .725 moles of NaOH since the molar mass of NaOH is 39.997. We do the calculation 29/39.997 to find the number of moles.

The unit M or molarity represents moles per liter. .171 M NaOH means there are .171 moles of NaOH in one liter of the solution. To find the volume needed, we can divided the number of moles we have which is .725 by .171 to find liters. This gives us 4.240 L or 4240.08 ml.

If you are confused by the calculation, you can look at how the units change: M=Mol/L .725 is in moles. We do mol/(mol/L) which is the same as mol*(L/mol) so the mol cancels and we are left with a unit in liters which we can easily convert to milliliters.

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