Question

Calculate the molality of each of the following solutions: (a) $8.66\mathrm{g}$ benzene $\left({\mathrm{C}}_6{\mathrm{H}}_6\right)$ dissolved in $23.6\mathrm{g}$ carbon tetrachloride $\left({\mathrm{CCl}}_4\right)$, (b) $4.80\mathrm{g}\mathrm{NaCl}$ dissolved in $0.350\mathrm{L}$ of water.

Short answer

The molality of benzene in carbon tetrachloride is 4.69 mol/kg, and the molality of sodium chloride in water is 0.2349 mol/kg.

Step-by-step solution

Determine the molar mass of benzene

To determine the molar mass of benzene, sum the molar mass of each constituent atom in the benzene molecule (C6H6): $6\mathrm{C}+6\mathrm{H}$ \ Molar Mass of Carbon (C): 12.01 g/mol Molar Mass of Hydrogen (H): 1.01 g/mol Molar mass of benzene: $6\times 12.01+6\times 1.01$ = 72.06 + 6.06 = 78.12 g/mol

Calculate the moles of benzene

Divide the mass of benzene by its molar mass: moles of benzene = ${\displaystyle \frac{8.66g}{78.12g/mol}}$ = 0.1108 mol

Convert the mass of carbon tetrachloride to kilograms

Since 1000 g = 1 kg, divide the mass of carbon tetrachloride by 1000 to convert it to kilograms: $23.6g={\displaystyle \frac{23.6}{1000}}kg=0.0236kg$

Calculate molality

Remember that molality = ${\displaystyle \frac{molesofsolute}{kgofsolvent}}$ molality = ${\displaystyle \frac{0.1108mol}{0.0236kg}}$ = 4.69 mol/kg Thus, the molality of benzene in carbon tetrachloride is 4.69 mol/kg. **Solution (b):** Sodium Chloride in Water

Determine molar mass of sodium chloride

Sodium chloride (NaCl) has the following molar masses: Molar Mass of Sodium (Na): 22.99 g/mol Molar Mass of Chlorine (Cl): 35.45 g/mol Molar mass of sodium chloride: 22.99 + 35.45 = 58.44 g/mol

Calculate the moles of sodium chloride

Divide the mass of sodium chloride by its molar mass: moles of sodium chloride = ${\displaystyle \frac{4.80g}{58.44g/mol}}$ = 0.0822 mol

Convert the volume of water to mass, then to kilograms

The density of water is approximately 1 g/mL or 1 g/cm³. Since 1000 mL = 1 L, we have: $Density\times Volume=Mass$ $1{\displaystyle \frac{g}{mL}}\times 350mL=350g=0.350kg$

Calculate molality

Remember that molality = ${\displaystyle \frac{molesofsolute}{kgofsolvent}}$ molality = ${\displaystyle \frac{0.0822mol}{0.350kg}}$ = 0.2349 mol/kg Thus, the molality of Sodium Chloride in water is 0.2349 mol/kg.