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"
