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[tex]\tt{\huge{\red{Solution:}}}[/tex]
Step 1: Calculate the number of moles of solute (KBr).
Note that the molar mass of KBr is 119.00 g/mol.
[tex]\begin{aligned} \text{moles of solute} & = \frac{\text{mass of solute}}{\text{molar mass of solute}} \\ & = \frac{\text{40.0 g}}{\text{119.00 g/mol}} \\ & = \text{0.336 mol} \end{aligned}[/tex]
Step 2: Calculate the mass of solvent (water).
Note that the density of water is 1 g/mL.
[tex]\begin{aligned} \text{mass of solvent} & = \text{density of solvent} \times \text{volume of solvent} \\ & = \text{1 g/mL} \times \text{800.0 mL} \\ & = \text{800.0 g} \\ & = 800.0 g \: \cancel{\text{g}} \times \frac{\text{1 kg}}{1000 \: \cancel{\text{g}}} \\ & = \text{0.8000 kg} \end{aligned}[/tex]
Final Step: Calculate the molality of solution.
Note that 1 molal (m) is equal to 1 mol/kg.
[tex]\begin{aligned} \text{molality} & = \frac{\text{moles of solute}}{\text{mass of solvent}} \\ & = \frac{\text{0.336 mol}}{\text{0.8000 kg}} \\ & = \text{0.420 mol/kg} \\ & = \boxed{0.420 \: m} \end{aligned}[/tex]
Hence, the molality of the solution is 0.420 m.
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