# What is the molality of a solution prepared by dissolving 150.0 g C_6H_12O_6 in 600.0 g H_2O?

Jul 23, 2016

The molality of the solution is 1.388 m.

#### Explanation:

Molality $\left(\text{m}\right)$ is a way of indicating the concentration of a solute dissolved in a solution. It is calculated by dividing the moles of solute by the mass of the solvent in kg.

"molality"=("moles solute")/("kg solvent")

You have been given the masses of both the solute and the the solvent in grams. You will need to convert the mass of ${\text{C"_6"H"_12"O}}_{6}$ to moles using its molar mass, and you will need to convert the mass of water from grams to kilograms.

Molar mass of ${\text{C"_6"H"_12"O}}_{6}$ is $\text{180.15588 g/mol}$.
https://pubchem.ncbi.nlm.nih.gov/compound/5793section=Top

Determine moles of $\text{C"_6"H"_12"O"_6}$:

"mol solute"=150.0"g C"_6"H"_12"O"_6xx(1"mol C"_6"H"_12"O"_6)/(180.15588"g C"_6"H"_12"O"_6)=color(blue)("0.8326 mol C"_6"H"_12"O"_6")

Convert mass of solvent from g to kg.

600.0"g H"_2"O"xx(1"kg H"_2"O")/(1000"g H"_2"O")=color(red)("0.6000 kg H"_2"O")

Calculate molality:

"molality"=("moles solute")/("kg solvent")

"molality"=(color(blue)(0.8326"mol"))/(color(red)(0.6000"kg"))=color(purple)("1.388 mol/kg=1.388 m")#