Final answer:
There are 6.0 grams of sodium hydroxide in a 250.0 ml of a 0.600M NaOH solution, calculated by first converting the volume to liters, determining the moles using molarity, and then converting moles to grams using the molar mass of NaOH.
Step-by-step explanation:
To calculate the mass of sodium hydroxide (NaOH) present in a 250.0 ml of a 0.600M NaOH solution, we first need to convert the volume from milliliters to liters since molarity (M) is expressed in moles per liter (mol/L). To do this, we divide the volume by 1000, converting 250.0 ml to 0.250 liters.
Now that the quantities are in proper units, we use the definition of molarity which is moles of solute per liter of solution:
Moles of NaOH = Molarity (M) × Volume (L) = 0.600M × 0.250L = 0.150 moles NaOH.
The molar mass of NaOH is approximately 40 g/mol, so we can find the mass by multiplying the number of moles by the molar mass:
Mass of NaOH = Moles of NaOH × Molar mass of NaOH = 0.150 moles × 40 g/mol = 6.0 grams.
Therefore, there are 6.0 grams of sodium hydroxide in a 250.0 ml of a 0.600M NaOH solution.