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