Calculate the mole fraction of a mixture of NaBr and KBr which is 0.981g of mass, given that we are treating the mixture with excessive AgNO3. 1.690g of AgBr is formed.
This is two unknowns so it requires two equations and solve them simultaneously. First, you must determine the grams NaBr and grams KBr.
Let X = grams NaBr
and Y = grams KBr
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equation 1 is X + Y = 0.981
equation 2 is obtained by converting grams X to grams AgBr and converting grams Y to grams AgBr. You know the total grams AgBr is 1.690. Equation 2 is
(mm stands for molar mass)
X(mmAgBr/mmNaBr)+(Y(mmAgBr/mmKBr) = 1.690.
Solve those two equations simultaneously for X and Y and that gives you grams NaBr and grams KBr.
mols NaBr = grams/molar mass = ?
mols KBr = grams/molar mass = ?
XNaBr = mols NaBr/total mols
XKBr = mols KBr/total mols.
Post your work if you stuck.
To calculate the mole fraction of a component in a mixture, we first need to determine the moles of each component present.
Given:
Mass of the mixture (NaBr + KBr) = 0.981g
Mass of AgBr formed = 1.690g
Step 1: Determine the moles of AgBr formed.
First, we need to calculate the molar mass of AgBr (silver bromide) using the atomic masses of silver (Ag) and bromine (Br).
Ag: atomic mass = 107.87 g/mol
Br: atomic mass = 79.90 g/mol
Molar mass of AgBr = (Ag atomic mass) + (Br atomic mass) = 107.87 g/mol + 79.90 g/mol = 187.77 g/mol
To find the moles of AgBr, divide its mass by its molar mass:
Moles of AgBr = Mass of AgBr formed / Molar mass of AgBr
Moles of AgBr = 1.690g / 187.77 g/mol
Step 2: Determine the moles of NaBr and KBr in the mixture.
Since we treated the mixture with excess AgNO3, all of the bromine in the NaBr and KBr reacts to form AgBr. Therefore, the moles of AgBr formed is equal to the total moles of bromine in NaBr and KBr.
To find the moles of NaBr and KBr, we can use their molar masses:
Na: atomic mass = 22.99 g/mol
Br: atomic mass = 79.90 g/mol
K: atomic mass = 39.10 g/mol
Molar mass of NaBr = (Na atomic mass) + (Br atomic mass) = 22.99 g/mol + 79.90 g/mol = 102.89 g/mol
Molar mass of KBr = (K atomic mass) + (Br atomic mass) = 39.10 g/mol + 79.90 g/mol = 119 g/mol
Let's assume the number of moles of NaBr is "x," then the number of moles of KBr is "0.981 - x" since the total moles of the mixture is 0.981 g.
Using the equation:
Moles of AgBr = Moles of NaBr + Moles of KBr
We can write:
1.690g / 187.77 g/mol = (x moles / 102.89 g/mol) + ((0.981 - x) moles / 119 g/mol)
Now, solve this equation to find the value of x, which represents the moles of NaBr:
1.690 / 187.77 = x / 102.89 + (0.981 - x) / 119
Solving this equation will give the value of x, representing the moles of NaBr.
Step 3: Calculate the mole fraction of NaBr in the mixture.
Mole fraction of NaBr = Moles of NaBr / Total moles of the mixture.
Total moles of the mixture = Moles of NaBr + Moles of KBr.
Using the calculated values, you can now determine the mole fraction of NaBr in the mixture.