Solubility of lead(ii)trioxonitrate is 1000g/dm3 per 1000g of h2O at 85 degrees and 5000g of h2O. Calculate the mass of lead(ii)trioxonitrate will crystalise if 120g of the saturated solutionis cooled to 40 degrees
You need to rewrite this question. As it is it is garbled and makes no sense.
BTW, the IUPAC name for the weird compound you have named is lead nitrate.
To solve this problem, we need to find the amount of lead(II) trioxonitrate that will crystallize when the temperature is lowered from 85 degrees to 40 degrees.
Step 1: Calculate the mass of lead(II) trioxonitrate in the saturated solution at 85 degrees:
Given:
Solubility = 1000g/dm3 per 1000g of H2O
Volume of water = 5000g
Using the solubility, we can calculate the mass of lead(II) trioxonitrate in the saturated solution at 85 degrees:
Mass of lead(II) trioxonitrate = Solubility x Volume of water
Mass of lead(II) trioxonitrate = 1000g/dm3 x (5000g/1000g)
Mass of lead(II) trioxonitrate = 5000g
Step 2: Calculate the mass of lead(II) trioxonitrate that will crystallize at 40 degrees:
Given:
Mass of saturated solution = 120g
Temperature change = 85 degrees to 40 degrees
Since the solubility changes with temperature, we need to find the solubility of lead(II) trioxonitrate at 40 degrees.
Step 3: Calculate the new solubility at 40 degrees using a linear approximation:
Given:
Solubility at 85 degrees = 1000g/dm3
Solubility at 40 degrees = Unknown
We can use the formula:
Solubility at 40 degrees = Solubility at 85 degrees x (40 degrees / 85 degrees)
Solubility at 40 degrees = 1000g/dm3 x (40 degrees / 85 degrees)
Solubility at 40 degrees ≈ 470.59g/dm3
Step 4: Calculate the mass of lead(II) trioxonitrate that will crystallize at 40 degrees:
Mass of lead(II) trioxonitrate = Mass of saturated solution - Mass of lead(II) trioxonitrate at 40 degrees
Mass of lead(II) trioxonitrate = 120g - (470.59g/dm3 x (120g/1000g))
Mass of lead(II) trioxonitrate = 120g - 56.47g
Mass of lead(II) trioxonitrate ≈ 63.53g
Therefore, approximately 63.53 grams of lead(II) trioxonitrate will crystallize when 120 grams of the saturated solution is cooled from 85 degrees to 40 degrees.
To solve this problem, we need to use the concept of solubility and the given data. Here are the steps to calculate the mass of lead(II) trioxonitrate that will crystallize:
Step 1: Determine the solubility of lead(II) trioxonitrate at 85 degrees Celsius.
According to the given information, the solubility of lead(II) trioxonitrate is 1000g/dm3 (grams per cubic decimeter) per 1000g of water (H2O) at 85 degrees Celsius. This means that in 1000g of water, you can dissolve 1000g of lead(II) trioxonitrate.
Step 2: Calculate the amount of lead(II) trioxonitrate dissolved in 120g of the saturated solution.
Since we know the solubility is 1000g/dm3, we can calculate the amount of lead(II) trioxonitrate present in 120g of the saturated solution using the formula:
Mass of solid = (Mass of solution / Volume of solution) x Solubility
First, we need to find the volume of the solution. We are given that there is 120g of solution. The density of water is approximately 1g/cm3 or 1g/mL. Therefore, the volume of the solution is 120 mL (since the weight and volume of water are the same in this case).
Now, applying the formula:
Mass of solid = (120g / 120 mL) x 1000g/dm3
Mass of solid = (1g/mL) x 1000g/dm3
Mass of solid = 1000g
So, 120g of the saturated solution contains 1000g of lead(II) trioxonitrate.
Step 3: Calculate the mass of lead(II) trioxonitrate that will crystallize when the solution is cooled to 40 degrees Celsius.
Since the solubility decreases with decreasing temperature, some of the lead(II) trioxonitrate will crystallize out. We need to find out how much mass will precipitate.
To calculate this, we need to determine the solubility of lead(II) trioxonitrate at 40 degrees Celsius. Unfortunately, the solubility at this temperature is not given. Therefore, we will assume that the solubility at 40 degrees Celsius is the same as at 85 degrees Celsius (1000g/dm3).
Now, we can calculate the mass of lead(II) trioxonitrate that will crystallize out using the formula:
Mass of solid = (Mass of solution / Volume of solution) x (Solubility at 40 degrees Celsius)
Again, the volume of the solution remains the same, which is 120 mL.
Mass of solid = (120g / 120 mL) x 1000g/dm3
Mass of solid = (1g/mL) x 1000g/dm3
Mass of solid = 1000g
Therefore, 1000g of lead(II) trioxonitrate will crystallize out when the saturated solution is cooled to 40 degrees Celsius.
In summary, when 120g of the saturated solution of lead(II) trioxonitrate at 85 degrees Celsius is cooled to 40 degrees Celsius, 1000g of the lead(II) trioxonitrate will crystallize out.