What mass of calcium bromide is needed to prepare 150.0 mL of a 3.50 M solution? (Assume that the molecular weight of CaBr2 is 200.618 g/mol)
A. 105 g
B. 100.42 g
C. 0.105 g
D. 103.35 g
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To calculate the mass of calcium bromide needed to prepare the solution, we can use the formula:
Mass (g) = volume (L) x molarity (mol/L) x molar mass (g/mol)
First, we need to convert the volume from mL to L:
150.0 mL = 150.0 mL x (1 L / 1000 mL) = 0.150 L
Next, we can substitute the given values into the formula:
Mass (g) = 0.150 L x 3.50 mol/L x 200.618 g/mol
Now, let's calculate the mass:
Mass (g) = 10.5071 g
Rounded to the correct number of significant figures, the mass of calcium bromide needed is:
Mass (g) = 10.51 g
The correct answer is not listed among the answer choices.
To determine the mass of calcium bromide needed to prepare the solution, we need to use the formula:
M1V1 = M2V2
Where:
M1 = initial molarity of the solution (unknown)
V1 = initial volume of the solution (unknown)
M2 = final molarity of the solution (3.50 M)
V2 = final volume of the solution (150.0 mL)
First, rearrange the formula to solve for M1:
M1 = (M2 * V2) / V1
Now, plug in the known values:
M1 = (3.50 M * 150.0 mL) / V1
Since we are trying to find the mass of calcium bromide, we need to convert the volume of the solution into liters. There are 1000 mL in 1 L, so:
V1 = 150.0 mL / 1000 mL/L = 0.150 L
Now, plug in the value for V1:
M1 = (3.50 M * 0.150 L) / V1
Next, we need to convert the molarity of the solution to moles per liter. The molecular weight of CaBr2 is given as 200.618 g/mol. So, for every 1 mole of CaBr2, there are 200.618 grams. The molar mass can be used as a conversion factor to convert moles to grams or grams to moles.
Moles of CaBr2 = M1 * V1
Now, plug in the values:
Moles of CaBr2 = (3.50 M * 0.150 L) / V1
Now, we need to convert moles of CaBr2 to grams using the molar mass:
Mass of CaBr2 = Moles of CaBr2 * molar mass
Mass of CaBr2 = [(3.50 M * 0.150 L) / V1] * molar mass
Mass of CaBr2 = [(3.50 M * 0.150 L) / V1] * 200.618 g/mol
At this point, we do not have a specific value for V1, but we know that it should be a value that gives us one of the given answer choices. Let's try the values one by one.
A. 105 g:
V1 = (3.50 M * 0.150 L) / (105 g / (200.618 g/mol))
To find V1, divide the mass of CaBr2 by the molar mass and plug in the values:
V1 = (3.50 M * 0.150 L) / (105 g / (200.618 g/mol))
V1 = 0.5301 L
Since V1 is not equal to 0.150 L, which is the given final volume, option A is not correct.
Repeat the process for the other answer choices to see if any of them result in V1 being equal to the final volume of the solution (0.150 L). The correct answer, therefore, is the one that gives V1 as 0.150 L.