We're asked to find the number of *moles* of calcium nitrate are in #75# #"mL"# of a #0.25# molar solution of #"Ca(NO"_3")"_2#.

To solve this equation, we can use the **molarity equation**:

#"molarity" = "moles solute"/"liters solution"#

We need the solution's volume to be in #"L"#, so let's convert it:

#75cancel("mL")((1color(white)(l)"L")/(10^3cancel("mL"))) = 0.075# #"L"#

We're asked to find the number of moles, so

#"moles solute" = ("molarity")("liters solution")#

#"mol Ca(NO"_3")"_2 = (0.25"mol"/(cancel("L")))(0.075cancel("L"))#

# = color(red)(0.019# #color(red)("mol Ca(NO"_3")"_2#

rounded to two sig figs.

Therefore, there are #color(red)(0.019# #sfcolor(red)("moles of calcium nitrate"# in #75# #"mL"# of a #0.25M# solution.