The freezing point is -1.19 °C.
The formula for calculating freezing point depression #ΔT_"f"# is
#color(blue)(bar(ul(|color(white)(a/a)ΔT_"f" = iK_fbcolor(white)(a/a)|)))" "#
where
#i# is the van't Hoff factor
#K_"f"# is the molal freezing point depression constant
#"b"# is the molality of the solution.
In this problem,
#i= 2#, because 1 mol of #"NaCl"# gives 2 mol of particles
#K_text(f) = "1.86 °C·kg·mol"^"-1"#
#b = "0.321 mol·kg"^"-1"#
∴ #ΔT_"f" = 2 × "1.86 °C"·color(red)(cancel(color(black)("kg·mol"^"-1"))) × 0.321 color(red)(cancel(color(black)("mol·kg"^"-1"))) = "1.194 °C"#
#T_text(f) = T_text(f)^° - ΔT_text(f) = "0 °C - 1.194 °C" = "-1.19 °C"#
Note: The answer can have only 3 significant figures (two decimal places), because that is all you gave for the molality of the solution.
