Determine the mole fraction of lithium bromate in a 6.32 M aqueous solution of lithium bromate. The density of the solution is 1.16 g mL-1.
I understand that 6.32 M of solution= mols of the compound.
I multiplied 1000mL by the density to get grams. I don't know what to do next.
To determine the mole fraction of lithium bromate in the solution, you need to calculate the moles of lithium bromate in the solution first. You correctly identified that 6.32 M of solution corresponds to the moles of the compound.
To calculate the moles of lithium bromate, you can use the formula:
moles = concentration (molarity) × volume (in liters)
Given that the concentration is 6.32 M, you need to convert the volume of the solution to liters. Here's how you can do it:
1. Calculate the volume of the solution in milliliters (mL):
volume = mass / density
Given the density is 1.16 g/mL, you can use the mass as 1000 mL (1 L) since the density is given in grams per milliliter. This gives you:
volume = 1000 mL
2. Convert the volume to liters:
volume = 1000 mL × (1 L / 1000 mL) = 1 L
3. Calculate the moles of lithium bromate:
moles = concentration × volume
= 6.32 M × 1 L
= 6.32 moles
Now that you have the moles of lithium bromate, you can calculate the mole fraction. The mole fraction (χ) is calculated as the moles of solute divided by the total moles of all components in the solution.
Since there is only lithium bromate as the solute in the solution, the mole fraction of lithium bromate (χLiBrO3) is:
χLiBrO3 = moles of LiBrO3 / (moles of LiBrO3 + moles of water)
Since there is no information given about the amount of water present in the solution, we assume it is a typical dilute aqueous solution, which means the mole fraction of water can be considered negligible compared to the moles of lithium bromate.
Therefore, you can approximate the mole fraction as:
χLiBrO3 ≈ moles of LiBrO3 / (moles of LiBrO3)
Simplifying this further, χLiBrO3 is approximately equal to 1.
So, the mole fraction of lithium bromate in the given solution is 1.
To determine the mole fraction of lithium bromate in the aqueous solution, you need to first convert the mass of the solution to moles of lithium bromate.
Step 1: Calculate the mass of the solution
Given that the density of the solution is 1.16 g/mL and the volume is 1000 mL, you can calculate the mass of the solution as follows:
Mass of solution = density x volume = 1.16 g/mL x 1000 mL = 1160 g
Step 2: Calculate the moles of lithium bromate
To find out the moles of lithium bromate, you need to know the molar mass of lithium bromate (LiBrO3). The molar mass can be calculated by adding the atomic masses of each individual element in the compound. In this case, lithium (Li) has an atomic mass of 6.94 g/mol, bromine (Br) has an atomic mass of 79.90 g/mol, and oxygen (O) has an atomic mass of 16.00 g/mol.
Molar mass of LiBrO3 = (6.94 g/mol x 1) + (79.90 g/mol x 1) + (16.00 g/mol x 3) = 109.84 g/mol
Now you can calculate the moles of lithium bromate using the following formula:
Moles of compound = Mass of compound / Molar mass
Moles of lithium bromate = 1160 g / 109.84 g/mol
Step 3: Calculate the mole fraction
Mole fraction = Moles of lithium bromate / Total moles of all components
In this case, the aqueous solution only contains lithium bromate, so the total moles of all components would be equal to the moles of lithium bromate.
Mole fraction of lithium bromate = Moles of lithium bromate / Moles of lithium bromate
Therefore, the mole fraction of lithium bromate in the 6.32 M aqueous solution of lithium bromate can be calculated by dividing the moles of lithium bromate (obtained in Step 2) by itself.