What volume of 0.716M KBr solution is needed to provide 10.5g of kBr?
How many mols is 10.5g KBr?
mols = grams/molar mass.
Then M KBr = mols KBr/L KBr. You know M and mols, solve for L.
Need answer
To calculate the volume of a 0.716M KBr solution needed to provide 10.5g of KBr, we first need to determine the number of moles of KBr.
Step 1: Calculate the molar mass of KBr:
KBr: 1 atom of K (39.10 g/mol) + 1 atom of Br (79.90 g/mol)
= 39.10 g/mol + 79.90 g/mol
= 119.0 g/mol
Step 2: Convert the mass of KBr to moles:
Number of moles = mass / molar mass
Number of moles = 10.5 g / 119.0 g/mol
Number of moles ≈ 0.0882 mol
Step 3: Use the formula:
Molarity = moles of solute / volume of solution (in liters)
To find the volume of the solution, we rearrange the formula:
Volume of solution (in liters) = moles of solute / molarity
Step 4: Calculate the volume of the solution in liters:
Volume of solution = 0.0882 mol / 0.716 M
Volume of solution ≈ 0.123 L
Step 5: Convert the volume to milliliters:
1 L = 1000 mL
0.123 L ≈ 123 mL
Therefore, approximately 123 mL of the 0.716M KBr solution is needed to provide 10.5g of KBr.
To find the volume of the 0.716M KBr solution needed to provide 10.5g of KBr, you need to use the formula:
Volume (in liters) = Amount of substance (in moles) / Concentration (in moles/liter)
First, we need to calculate the moles of KBr (potassium bromide) from the given mass. To do this, we use the formula:
Moles = Mass / Molar Mass
The molar mass of KBr is found by adding the atomic masses of potassium (K) and bromine (Br):
Molar Mass of K = 39.1 g/mol
Molar Mass of Br = 79.9 g/mol
Molar Mass of KBr = Molar Mass of K + Molar Mass of Br = 39.1 g/mol + 79.9 g/mol = 119 g/mol
Now, substitute the given mass into the moles formula:
Moles = 10.5g / 119 g/mol = 0.0882 mol
Finally, using the volume formula, we can find the volume of the solution:
Volume = Moles / Concentration
Given concentration = 0.716M
Volume = 0.0882 mol / 0.716 mol/L = 0.123 L or 123 mL
Therefore, you would need 123 mL of the 0.716M KBr solution to provide 10.5g of KBr.