How many grams of sodium hydroxide are present in 247.0 mL of a 0.300 M NaOH solution?

The grams of NaOH that are present in 247.0 ml of a 0.300M NaOH solution is 2.964 grams

calculation

Step 1: find moles of NaOH

moles = molarity x volume in liters

volume in liters = 247.0 / 1000 = 0.247 liters

molarity = 0.300 M = 0.300 mol/l

moles = 0.300 mol/l x 0.247 = 0.0741 moles

step 2: find mass of NaOH

mass = moles x molar mass

The molar mass of NaOH = 23 + 1 + 16 = 40 g/mol

mass=0.0741 moles x40 g/mol = 2.964 grams