What mass of lithium iodide is needed to make 0.35 m solution with 1.4kg of solvent?
The correct answer and explanation is:
To find the mass of lithium iodide (LiI) needed to make a 0.35 molal (m) solution, we can use the formula for molality: m=moles of solutemass of solvent in kgm = \frac{\text{moles of solute}}{\text{mass of solvent in kg}}
Where:
- m is the molality (mol/kg),
- moles of solute is the number of moles of lithium iodide (LiI),
- mass of solvent in kg is the mass of the solvent, which is 1.4 kg in this case.
Step 1: Calculate the number of moles of LiI required
Rearrange the molality formula to find the moles of solute: moles of LiI=m×mass of solvent in kg\text{moles of LiI} = m \times \text{mass of solvent in kg}
Substitute the known values: moles of LiI=0.35 mol/kg×1.4 kg=0.49 mol\text{moles of LiI} = 0.35 \, \text{mol/kg} \times 1.4 \, \text{kg} = 0.49 \, \text{mol}
Step 2: Calculate the mass of LiI
To calculate the mass of LiI required, we need to use the molar mass of LiI. The molar mass of lithium iodide is calculated by adding the atomic masses of lithium (Li) and iodine (I):
- Atomic mass of Li = 6.94 g/mol,
- Atomic mass of I = 126.90 g/mol.
Molar mass of LiI=6.94 g/mol+126.90 g/mol=133.84 g/mol\text{Molar mass of LiI} = 6.94 \, \text{g/mol} + 126.90 \, \text{g/mol} = 133.84 \, \text{g/mol}
Now, calculate the mass of LiI needed: mass of LiI=moles of LiI×molar mass of LiI\text{mass of LiI} = \text{moles of LiI} \times \text{molar mass of LiI} mass of LiI=0.49 mol×133.84 g/mol=65.43 g\text{mass of LiI} = 0.49 \, \text{mol} \times 133.84 \, \text{g/mol} = 65.43 \, \text{g}
Final Answer:
The mass of lithium iodide required to make a 0.35 molal solution with 1.4 kg of solvent is 65.43 g.
This result is based on the definition of molality, which measures the amount of solute (in moles) per kilogram of solvent. Understanding how to calculate molality is crucial when preparing solutions for various chemical processes, ensuring the correct concentration for reactions or analyses.