Question #4ae2d

m/v %

The formula for mass/volume percent is

`color(blue)(bar(ul(|color(white)(a/a)"m/v %" = "mass of solute (g)"/"volume of solution (mL)" × 100 %color(white)(a/a)|)))" "`

`"m/v %" = "75.0 g"/"3500 mL" × 100 % = 2.1 %`

Molarity

The formula for molarity is

`color(blue)(bar(ul(|color(white)(a/a)"Molarity" = "moles"/"litres"color(white)(a/a)|)))" "`

or

`c = n/V`

The molar mass of glucose (`"C"_6"H"_12"O"_6`) is 180.16 g/mol.

∴ `n = 75.0 color(red)(cancel(color(black)("g C"_6"H"_12"O"_6))) × ("1 mol C"_6"H"_12"O"_6)/(180.16 color(red)(cancel(color(black)("g C"_6"H"_12"O"_6)))) = "0.4163 mol C"_6"H"_12"O"_6`

`c = "0.4163 mol"/"3.5 L" = "0.12 mol/L"`

Osmolality

The formula for molality is

`color(blue)(bar(ul(|color(white)(a/a)"Molality" = "moles of solute"/"kilograms of solvent"color(white)(a/a)|)))" "`

The density of the solution is slightly greater than that of pure water because of the dissolved sugar.

You don’t give a value for the density, so I shall assume a value of 1.008 g/mL.

The mass of 1 L of solution is

`"Mass" = 1000 color(red)(cancel(color(black)("mL"))) × "1.008 g"/(1 color(red)(cancel(color(black)("mL")))) = "1008 g"`

We know that 1 L of solution contains 21 g or 0.12 mol of glucose, so

`"Mass of water" = "1008 g - 22 g" = "986 g" = "0.986 kg"`

∴ The molality `b` is

`b = "moles"/"kilograms" = "0.12 mol"/"0.986 kg" = "0.12 mol/kg"`

An osmole is the number of moles solute particles of solute that contribute to the osmotic pressure of a solution.

Osmolality is the number of osmoles in 1 kg of solvent.

Singe glucose does not dissociate in solution, the number of osmoles is the same as the number of moles.

`"Osmolality" = (0.12 color(red)(cancel(color(black)("mol"))))/"1 kg" × "1 osmol"/(1 color(red)(cancel(color(black)("mol")))) = "0.12 osmol/kg"`