Al2(s04)3 solution of 1 molal concentration is present in 1 litre solution of 2.684g/cc.How many moles of bas04 would be precipitation on adding bacl2 in excess?
The answer 6M
I do not understand pls explain step by step
Thank a lot all tutors
I agree the problem makes no sense. I have brought this up before to Ken; m stands for molal while M sands for molar.
To solve this question, we need to use the concept of stoichiometry and the molar ratios from the balanced chemical equation.
Step 1: Determine the number of moles of Al2(SO4)3 in 1 liter of the solution.
To find the number of moles, we need to know the molar mass of Al2(SO4)3. The atomic masses of Al, S, and O are 27, 32, and 16, respectively.
Molar mass of Al2(SO4)3 = 2 * (27) + 3 * (32 + 4 * 16) = 342 g/mol
Given that the solution is 1 molal, which means there is 1 mole of solute in 1 kg of solvent (water). Since 1 liter of water weighs approximately 1 kg, we have 1 mole of Al2(SO4)3 in 1 liter of the solution.
Step 2: Determine the number of moles of BaCl2 required for complete reaction with Al2(SO4)3.
From the balanced chemical equation, we can see that the stoichiometric ratio between Al2(SO4)3 and BaCl2 is 1:3.
Al2(SO4)3 + 3BaCl2 -> 2AlCl3 + 3BaSO4
Since there is an excess of BaCl2, we need to calculate the number of moles of BaCl2 required for complete reaction with 1 mole of Al2(SO4)3. It can be calculated as follows:
Moles of BaCl2 = 3 * Moles of Al2(SO4)3
Step 3: Convert the number of moles of BaCl2 to molar concentration.
The solution density is given as 2.684 g/cc, which means 1 liter of the solution weighs 2684 grams.
To calculate the molar concentration (M) of BaCl2, we need to convert the mass of BaCl2 to moles:
Moles of BaCl2 = (Mass of BaCl2) / (Molar mass of BaCl2)
Step 4: Calculate the concentration of BaCl2 in moles per liter (M).
Concentration (M) = (Moles of BaCl2) / (Volume of solution in liters)
Now you can substitute the values obtained from the calculations and solve for the concentration of BaCl2.
Based on the given answer of 6M, it seems that the concentration of BaCl2 required for the precipitation of BaSO4 is 6 moles per liter.
Step by step.
The first sentence. Weird. It says you have one M solution. Lets figure that.
1M=masspresent/(volume*molmass)
1M*molmass:*1Liter=mass
1*342.15gram=mass
or mass/cc=.342 g/cc
So why does it then state 2.684g/cc ? Nuts to this problem, your teacher is not thinking.
Finally, you aske how many moles of BaSO4 there would be, and you tell me the answer is 6 Molal? Nuts again.