We're asked to find the mass, in #"g"#, of #"NaOH"# in #248.0# #"mL"# of a #0.600M# #"NaOH soln"#.
To do this, we can use the molarity equation to calculate the number of moles of #"NaOH"#:
#"molarity" = "mol solute"/"L soln"#
The molarity is given as #color(green)(0.600M#, and the volume is #248.0# #"mL"#.
Since this volume must be in liters, we have
#248.0cancel("mL")((1color(white)(l)"L")/(10^3cancel("mL"))) = color(red)(0.2480# #color(red)("L"#
Let's find the moles of #"NaOH"#:
#"mol NaOH" = ("molarity")("L soln") = (color(green)(0.600M))(color(red)(0.2480color(white)(l)"L"))#
#= color(purple)(0.1488# #color(purple)("mol NaOH"#
Lastly, we'll use the molar mass of sodium hydroxide (#40.00# #"g/mol"#) to find the number of grams:
#color(purple)(0.1488)cancel(color(purple)("mol NaOH"))((40.00color(white)(l)"g NaOH")/(1cancel("mol NaOH"))) = color(blue)(ul(5.95color(white)(l)"g NaOH"#
