1.33dm³ of water at 70°c are saturated by 2.25moles of Pb(NO3)2, and 1.33dm³ of water at 18°c are saturated by 0.53mole of the same salt. If 4.50dm³ of the saturated solution are cooled from 70°c to 18°c.calculate the amount of solute that will be deposited in a). Moles b) grams. (Pb=207,N=14,O=16)
1.33dm3 of water at 70°C = 2.25moles of Pb(NO3)2 and 4.5dm3 of water contains x moles of the salt, by calculation, x=7.62 moles..likewise at 18°C, 1.33dm3 of water contains 0.53 moles of the same salt and at 4.5dm3 of water, it contains x..x is 4.5x0.53/1.33=1.79moles...so when the salt is cooled from 70°C to 18°C, the amount of salt deposited would be 7.62 - 1.79= 5.83moles..so 5.83 moles is the amount of salt that would be deposited when the saturated solution is cooled from 70°C to 18°C..then for b,no of mole=mass/molar mass..molar mass of the salt is 331, then the equivalent mass would be 5.83×331=1929.73g
This is extremely fantastic. I love it,but can you rexplain
How much Pb(NO3)2 is contained in the 4.5 dm^ at 70 C? That's
2.25 mols x 4.5/1.33 = approx 7.5 but you need a more accurate number.
Now you have 4.5 dm^3 containing about 7.5 mols Pb(NO3)2 at 70 C. You cool it to 18 C where the solubility is lower. How much can that 4.5 cm^3 hold at the lower temperature? That's
0.53 x 4.5/1.33 = about 1.8 or so; again you need a more accurate number. How much will come out of solution? It started with about 7.5 mols in solution. It ends with about 1.8 mols in solution. The difference comes out of solution or 7.5-1.8 = ? mols precipitated. Convert to grams for part b.
I don't understand how you got that x=7.62 pls explain more am very confused 🤔😕
I don't understand
Sure, I'll explain it in more detail.
First, we're given that 1.33 dm³ of water at 70°C is saturated with 2.25 moles of Pb(NO₃)₂. This means that the solution is holding the maximum amount of solute (Pb(NO₃)₂) that it can at that temperature and volume of water.
Next, we're asked to find out how much solute (Pb(NO₃)₂) can be held in 4.5 dm³ of water at 70°C. We can use the following formula:
moles of solute1/volume of solvent1 = moles of solute2/volume of solvent2
We know that moles of solute1 is 2.25 and volume of solvent1 is 1.33 dm³ (given in the problem). We're being asked to find moles of solute2, given that volume of solvent2 is 4.5 dm³. Plugging in the numbers, we get:
2.25 moles/1.33 dm³ = moles of solute2/4.5 dm³
Solving for moles of solute2, we get:
moles of solute2 = 2.25 moles/1.33 dm³ x 4.5 dm³ = 7.62 moles
So, at 70°C, 4.5 dm³ of water can hold a maximum of 7.62 moles of Pb(NO₃)₂.
Now, we need to find out how much solute (Pb(NO₃)₂) will be deposited when the solution is cooled from 70°C to 18°C. We're given that 1.33 dm³ of water at 18°C is saturated with 0.53 moles of Pb(NO₃)₂. Using the same formula as before, we can find out how much solute (Pb(NO₃)₂) can be held in 4.5 dm³ of water at 18°C:
0.53 moles/1.33 dm³ = moles of solute2/4.5 dm³
Solving for moles of solute2, we get:
moles of solute2 = 0.53 moles/1.33 dm³ x 4.5 dm³ = 1.79 moles
So, at 18°C, 4.5 dm³ of water can hold a maximum of 1.79 moles of Pb(NO₃)₂.
The solubility of Pb(NO₃)₂ decreases with decreasing temperature, so when the solution is cooled from 70°C to 18°C, some of the Pb(NO₃)₂ will come out of solution and be deposited as a solid. The amount of Pb(NO₃)₂ that comes out of solution is equal to the difference between the amount of Pb(NO₃)₂ in the solution at 70°C (7.62 moles) and the amount of Pb(NO₃)₂ that can be held in the solution at 18°C (1.79 moles):
amount of Pb(NO₃)₂ deposited = 7.62 moles - 1.79 moles = 5.83 moles
Finally, to find the mass of Pb(NO₃)₂ deposited, we can use the following formula:
mass = moles x molar mass
The molar mass of Pb(NO₃)₂ is 331 g/mol (207 g/mol for Pb, 14 g/mol for N, and 3 x 16 g/mol for O). Plugging in the numbers, we get:
mass = 5.83 moles x 331 g/mol = 1929.73 g
So the amount of Pb(NO₃)₂ that will be deposited when the saturated solution is cooled from 70°C to 18°C is 5.83 moles, or 1929.73 grams.
I understand now tnx
You're welcome! I'm glad I could help. If you have any more questions, don't hesitate to ask.
Ok NAME pls
My name is OpenAI, I'm an AI language model designed to assist with various tasks including answering questions and providing explanations.
Ok I'm Sophie
Nice to meet you Sophie! Let me know if there's anything else I can help you with.
On organic chemistry
Sure, I'd be happy to help with any organic chemistry questions you may have. What specifically do you need help with?
Isomers of nonane
Nonane has the molecular formula C9H20. Isomers are molecules that have the same molecular formula but different structural arrangements. Nonane can have multiple isomers due to the number of possible ways the carbon atoms can be arranged.
The first type of isomer is a structural isomer, which means that the carbon atoms are arranged differently in the molecule. Nonane has five structural isomers, which are:
1. Linear nonane: The carbon atoms are arranged in a straight line. This is the most common and simplest isomer of nonane.
2. Branched-chain nonane: This isomer has one branch point in the carbon chain.
3. 2,2-Dimethylhexane: This isomer has two methyl groups (CH3) attached to the second carbon atom in the chain.
4. 2,3-Dimethylhexane: This isomer has one methyl group attached to the second carbon atom in the chain and another methyl group attached to the third carbon atom.
5. 2,4-Dimethylhexane: This isomer has one methyl group attached to the second carbon atom in the chain and another methyl group attached to the fourth carbon atom.
In addition to the structural isomers, nonane also has stereoisomers, which are isomers that have the same structural formula but differ in the arrangement of their atoms in space. Nonane only has one type of stereoisomer, which is a cis-trans isomerism. However, it's important to note that cis-trans isomerism only occurs in compounds with double bonds and nonane does not have any double bonds.