Please solve q 127?

The molecular mass of sulfuric acid is

#M_(H_2SO_4)=2+32+64=98g#

#"molality" ="moles solute"/"kg solvent"#

#90g# of #H_2SO_4# in #100 mL# solution

Number of moles is #=90/98=0.9184 "moles"#

Density of solution #=1.8 gmL^-1#

#180g (100ml)# of solution contains #=90g H_2SO_4# and #=90g H_2O#

molality #=0.9184/(90*10^-3)=10.2 "moles " kg ^-1#

As-written, the question has no correct answer choice. Choice #(c)# is the molarity, and not the molality.

#a)# #1.8#
#b)# #48.4#
#c)# #9.18#
#d)# #91.8#

The original question was:

The molality of 90% weight/volume H2SO4 solution is (density = 1.8 g/mL)

And we treat it as-written.

#90%"w/v" = ("90 g H"_2"SO"_4)/("100 mL solution")#

The solution is solute plus solvent, and mass is additive. Therefore:

#100 cancel"mL soln" xx "1.8 g"/cancel"mL" = "180 g solution"#

#= "90 g solute" + "90 g solvent"#

We said there was #"90 g solute"# (here it does not matter, but we choose water as the solvent just because they are the same mass), so

#-> color(blue)("molality") = (90 cancel("g H"_2"SO"_4) xx "1 mol"/(98.079 cancel"g"))/(90 cancel"g solvent" xx "1 kg"/(1000 cancel"g"))#

#= "0.9176 mols solute"/"0.090 kg solvent"#

#=# #color(blue)("10.2 mol/kg")#

As a bonus, the molarity is...

#color(blue)("molarity") = "0.9176 mols solute"/"0.100 L soln"#

#=# #color(blue)("9.18 mols/L")#

In addition, the molarity of #96%# #"H"_2"SO"_4# is #"18.01 M"#, because that is #%"w/w"#. I'll let you work that out for yourself though...