We're asked to find the (molar) concentration of #"HBr"# in solution, given its mass percentage and density.

To do this, we'll first assume there is #1# #"L"# of solution, so the density is also written as

#1.5# #"g/mL"# #= 1500# #"g/L"#

We're given that #48%# of the mass (#1500# #"g"#) is #"HBr"#, so the mass of #"HBr"# is

#0.48·1500# #"g soln"# #= ul(720color(white)(l)"g HBr"#

Now, we use the **molar mass** of #"HBr"# (#80.912# #"g/mol"#) to calculate the number of *moles*:

#720cancel("g HBr")((1color(white)(l)"mol HBr")/(80.912cancel("g HBr"))) = color(red)(8.9# #color(red)("mol HBr"#

Molarity equation:

#"molarity" = "mol solute"/"L soln"#

#= (color(red)(8.9color(white)(l)"mol HBr"))/(underbrace(1color(white)(l)"L soln")_"we assumed 1 L") = color(blue)(8.9color(white)(l)"mol/L"#

Thus, option **(B)** is the correct answer.