The meaning of #5%# w/w is just, for example:
#"5 g"/"100 g"#
The definition of molality is:
#"mol solute"/"kg solvent"#
a) #5%# w/w glycerine (#"100. g = 0.100 kg"#)
In this case, glycerine is both the solute and the solvent since there is no other solvent. Let's say you had acquired exactly #"5.0000 g"# (my university has scales of that precision). The #"mol"#s of glycerine would be calculated to be:
#(5 cancel("g"))(("1 mol")/(92.0932 cancel("g"))) = "0.05429 mol"#
Therefore, your molality is:
#= "0.05429 mol glycerine"/"0.100 kg glycerine"#
#color(blue)("= 0.5429 m")# or "molal" of glycerine
b) #"10.0 g"# methanol in #"60.0 g (0.0600 kg)"# benzene
#"CH"_3"OH"#, methanol, has a molar mass of #"32.0416 g/mol"#
so the #"mol"#s of it turn out to be:
#(10.0 cancel("g"))("1 mol"/(32.0416 cancel("g"))) = "0.312 mol"#
The molality is therefore:
#= "0.312 mol methanol"/"0.0600 kg benzene"#
#color(blue)("= 5.20 m")# or "molal" of methanol
