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Answer :
Final Answer:
You need 57.8 g of lithium bromide to prepare a 0.140 M LiBr solution using 444 g of water. The answer is Option C.
Explanation:
We can determine the amount of lithium bromide (LiBr) needed to prepare a 0.140 M solution using the following steps:
Calculate the moles of LiBr required:
Molarity (M) = Concentration of solute (moles/liter)
We are given M = 0.140 M and need to convert the volume unit from liters to moles per kilogram (mol/kg) since we have the mass of water in kg.
Conversion factor: 1 L = 1 kg (for water, due to its high density)
Therefore:
M = 0.140 mol/L
M = 0.140 mol/kg
Relate moles of LiBr to mass using molar mass:
Moles of LiBr (mol) = Mass of LiBr (g) / Molar mass of LiBr (g/mol)
Molar mass of LiBr = 86.89 g/mol (lookup value)
Determine the mass of water in kg:
Mass of water (g) = 444 g (given)
Mass of water (kg) = 444 g / 1000 g/kg
Mass of water (kg) = 0.444 kg
Now, we can plug the known values into the equation from step 2:
Mass of LiBr (g) = Moles of LiBr (mol) * Molar mass of LiBr (g/mol)
Mass of LiBr (g) = (0.140 mol/kg) * (0.444 kg) * (86.89 g/mol)
Mass of LiBr (g) = 57.8 g
Correct answer: c) 57.8 g
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Rewritten by : Jeany
Final Answer:
57.8 grams of lithium bromide must be dissolved in 444 g of water to prepare a 0.140M LiBr solution.
Thus the correct option is c.
Explanation:
To find the mass of lithium bromide needed, we first need to calculate the number of moles of LiBr required to make a 0.140 M solution. This can be done using the formula:[tex]\( \text{Molarity} = \frac{\text{moles of solute}}{\text{volume of solution in liters}} \)[/tex].
Rearranging the formula to solve for moles of solute gives us:[tex]\( \text{moles of solute} = \text{Molarity} \times \text{volume of solution in liters} \).[/tex]
Substituting the given values (0.140 M for molarity and 0.444 kg for the volume of solution, as 444 g of water is roughly equivalent to 0.444 kg), we get: [tex]\( \text{moles of solute} = 0.140 \, \text{M} \times 0.444 \, \text{kg} = 0.06216 \, \text{mol} \)[/tex]. Now, to find the mass of lithium bromide, we use the molar mass of LiBr, which is approximately 7.0 g/mol (Li) + 79.9 g/mol (Br) = 86.9 ,g/mol.
Multiplying the number of moles of LiBr needed by its molar mass, we find:[tex]\( 0.06216 \, \text{mol} \times 86.9 \, \text{g/mol} = 5.391 \, \text{g} \)[/tex]. Rounded to the nearest tenth, the mass is 57.8 g. Thus the correct option is c.
