Question

Calculate the molality of a solution containing 275g glucose(`C_6H_12O_6`) dissolved in 1.08L of water? (Assume a density of 1.00 g/mL for water.)

Answer 1

The molality is 1.42 mol/kg, or 1.42 m.

Molality, m, is the moles of solute divided by the mass in kg of solvent.

Convert `1.08 "L"` of water to `"kg"` water.

`1.08"L"` x `(1.00 cancel"g")/(1 cancel"mL")xx(1 "kg")/(1000 cancel"g")xx(1000cancel"mL")/(1cancel"L")=1.08 "kg"`

Determine moles of 275 g glucose from its molar mass .
Molar mass of glucose is 180.1559 g/mol.
http://en.wikipedia.org/wiki/Glucose

`275 "g C"_6"H"_12"O"_6"` x `(1 "mol")/(180.1559 "g")=1.53 "mol C"_6"H"_12"O"_6"`

Determine Molality

`"m"=("mol")/("kg")=(1.53 "mol C"_6"H"_12"O"_6)/(1.08 "kg")=1.42 "mol/kg"`