How many milliliters of an aqueous solution of 0.238 M magnesium bromide is needed to obtain 9.01 grams of the salt?
L x M x molar mass = grams.
Substitute and solve for L, then convert to mL.
23233
To find the volume of the solution needed, you need to use the equation:
moles = mass / molar mass
First, calculate the molar mass of magnesium bromide (MgBr2):
Molar mass of MgBr2 = molar mass of Mg + 2 * molar mass of Br
The molar masses of Mg and Br are approximately 24.31 g/mol and 79.90 g/mol, respectively.
So, the molar mass of MgBr2 = 24.31 g/mol + 2 * 79.90 g/mol = 183.11 g/mol
Next, calculate the number of moles of MgBr2 in 9.01 grams using the equation:
moles = mass / molar mass
moles = 9.01 g / 183.11 g/mol
Now, use the molarity of the solution to calculate the volume of the solution in liters:
moles = molarity * volume (in liters)
Rearranging the equation gives:
volume (in liters) = moles / molarity
Substitute the calculated moles and given molarity into the equation to find the volume in liters.
Finally, convert the volume from liters to milliliters by multiplying by 1000, since there are 1000 milliliters in a liter.