(x-y)g of Al2(SO4)3 dissolved in 500cm^3 portion of a solution at 23°C to obtain a solubility of 2mol/dm^3. Also, (2x-y)g of acidified K2SO4 dissolved in 1000cm^3 portion of the solution in 78°C to obtain a solubility of 4mol/dm^3. Determine the value of (x+y)g. (Al=27, S=32, K=39, O=16, H=1).
Show workings bit by bit
Alright, let's solve this step by step:
1. First, let's calculate the number of moles of Al2(SO4)3 in the 500cm^3 portion of the solution.
Given that the solubility of Al2(SO4)3 is 2mol/dm^3, we can convert the volume from cm^3 to dm^3 by dividing by 1000:
Volume = 500cm^3 ÷ 1000 = 0.5dm^3
Number of moles of Al2(SO4)3 = solubility × volume
Number of moles = 2mol/dm^3 × 0.5dm^3
Number of moles = 1 mole
2. Now, let's calculate the mass of Al2(SO4)3 in the (x-y)g portion:
Given that the molar mass of Al2(SO4)3 is:
Al = 27g/mol
S = 32g/mol
O = 16g/mol
Molar mass of Al2(SO4)3 = (2 × Al) + (3 × S) + (12 × O)
Molar mass = (2 × 27g/mol) + (3 × 32g/mol) + (12 × 16g/mol)
Molar mass = 54g + 96g + 192g
Molar mass = 342g/mol
Mass = Number of moles × Molar mass
Mass = 1 mole × 342g/mol
Mass = 342g
So, the mass of (x-y)g of Al2(SO4)3 in the 500cm^3 portion is 342g.
3. Let's now calculate the number of moles of K2SO4 in the 1000cm^3 portion at 78°C.
Given that the solubility of K2SO4 is 4mol/dm^3, we can use the same process as before to calculate the number of moles:
Volume = 1000cm^3 ÷ 1000 = 1dm^3
Number of moles of K2SO4 = solubility × volume
Number of moles = 4mol/dm^3 × 1dm^3
Number of moles = 4 moles
4. Let's calculate the mass of K2SO4 in the (2x-y)g portion.
Given that the molar mass of K2SO4 is:
K = 39g/mol
S = 32g/mol
O = 16g/mol
Molar mass of K2SO4 = (2 × K) + S + (4 × O)
Molar mass = (2 × 39g/mol) + 32g/mol + (4 × 16g/mol)
Molar mass = 78g + 32g + 64g
Molar mass = 174g/mol
Mass = Number of moles × Molar mass
Mass = 4 moles × 174g/mol
Mass = 696g
So, the mass of (2x-y)g of acidified K2SO4 in the 1000cm^3 portion is 696g.
5. Now, let's set up an equation using the values we have calculated:
(x-y)g + (2x-y)g = 342g + 696g
Simplifying the equation gives us:
3x - 2y = 1038g
6. Finally, we can solve for the value of (x+y)g.
We can rearrange the equation we obtained in step 5:
3x = 1038g + 2y
x = (1038g + 2y)/3
Substituting this value of x into (x+y)g:
(x+y)g = [(1038g + 2y)/3] + y
(x+y)g = 1038g/3 + (2y/3) + y
(x+y)g = (1038g + 5y)/3
So, the value of (x+y)g is (1038g + 5y)/3.
To solve this problem, we need to use the concept of molar solubility and the molar mass of the compounds involved.
1. Let's start with Al2(SO4)3:
- The molar mass of Al2(SO4)3 is calculated as follows:
Al: 2 x 27 = 54 g/mol
S: 3 x 32 = 96 g/mol
O: 12 x 16 = 192 g/mol
Total molar mass = 54 + 96 + 192 = 342 g/mol
2. We are given the solubility of Al2(SO4)3 as 2 mol/dm^3.
- This means that for every 342 g of Al2(SO4)3, we have 2 moles.
3. We need to determine the mass required to achieve this solubility:
- The volume of the solution is given as 500 cm^3.
- First, we convert the volume from cm^3 to dm^3: 500 cm^3 / 1000 = 0.5 dm^3.
- Since the solubility is given in mol/dm^3, we have: (2 mol/dm^3) x (0.5 dm^3) = 1 mol.
4. To find the mass, we need to calculate the number of grams in 1 mol of Al2(SO4)3:
- 1 mol of Al2(SO4)3 weighs 342 g/mo;
- Therefore, the mass required to achieve a solubility of 2 mol/dm^3 is 342 g.
5. Now let's move on to acidified K2SO4:
- The molar mass of K2SO4 is calculated as follows:
K: 2 x 39 = 78 g/mol
S: 32 g/mol
O: 4 x 16 = 64 g/mol
Total molar mass = 78 + 32 + 64 = 174 g/mol.
6. We are given the solubility of K2SO4 as 4 mol/dm^3.
- This means that for every 174 g of K2SO4, we have 4 moles.
7. The volume of the solution is given as 1000 cm^3.
- We convert the volume from cm^3 to dm^3: 1000 cm^3 / 1000 = 1 dm^3.
- Since the solubility is given in mol/dm^3, we have: (4 mol/dm^3) x (1 dm^3) = 4 moles.
8. To find the mass, we need to calculate the number of grams in 4 moles of K2SO4:
- 4 moles of K2SO4 weighs 174 g/mol.
- Hence, the mass required to achieve a solubility of 4 mol/dm^3 is 4 x 174 g = 696 g.
9. Finally, we can determine the value of (x+y)g by setting up an equation:
- (x-y)g (Al2(SO4)3) = 342 g
- (2x-y)g (K2SO4) = 696 g
Let's solve these equations simultaneously:
From equation 1: x - y = 342
From equation 2: 2x - y = 696
Multiplying equation 1 by 2 gives us: (2x - 2y = 684)
Adding the results of equation 2 and the modified equation 1 will eliminate the 'y' term:
(2x - y) + (2x - 2y) = 696 + 684
4x - 3y = 1380
Solving the equation 4x - 3y = 1380 for 'x' and 'y,' we get:
4x = 1380 + 3y
x = (1380 + 3y) / 4
Substituting this value of 'x' into equation 1 gives us:
(1380 + 3y) / 4 - y = 342
Now we can solve for 'y':
Multiply equation (1380 + 3y) / 4 - y = 342 by 4 to remove the fraction:
1380 + 3y - 4y = 1368
Simplifying:
-y = 1368 - 1380
-y = -12
y = 12
Thus, substituting the value of 'y' into equation 1:
x - 12 = 342
x = 342 + 12
x = 354
Therefore, the values of (x+y)g = 354g + 12g = 366g.
To solve this problem, we can set up two equations based on the solubility of Al2(SO4)3 and K2SO4. Let's call the molar mass of Al2(SO4)3 M and the molar mass of K2SO4 N.
From the given information, we have:
(1) (x - y)g of Al2(SO4)3 dissolved in 500cm^3 of solution at 23°C to obtain a solubility of 2mol/dm^3.
(2) (2x - y)g of acidified K2SO4 dissolved in 1000cm^3 of solution at 78°C to obtain a solubility of 4mol/dm^3.
First, let's find the molar mass for Al2(SO4)3:
M = 2(Al atomic mass) + 3(S atomic mass) + 12(O atomic mass)
= 2(27 g/mol) + 3(32 g/mol) + 12(16 g/mol)
= 54 g/mol + 96 g/mol + 192 g/mol
= 342 g/mol
Next, let's find the molar mass for K2SO4:
N = 2(K atomic mass) + 1(S atomic mass) + 4(O atomic mass)
= 2(39 g/mol) + 1(32 g/mol) + 4(16 g/mol)
= 78 g/mol + 32 g/mol + 64 g/mol
= 174 g/mol
From equation (1), we know that the number of moles of Al2(SO4)3 dissolved is equal to:
(number of grams of Al2(SO4)3) / (molar mass of Al2(SO4)3) = (500 cm^3)(2 mol/dm^3)
Therefore, (x - y)g of Al2(SO4)3 is equal to:
(x - y)g / 342 g/mol = 500 cm^3 * (2 mol/dm^3)
Simplifying, we get:
(x - y)g = (500 cm^3 * 2 mol/dm^3) * 342 g/mol
(x - y)g = 1000 * 342 / 1000
(x - y)g = 342 g
From equation (2), we know that the number of moles of K2SO4 dissolved is equal to:
(number of grams of K2SO4) / (molar mass of K2SO4) = (1000 cm^3)(4 mol/dm^3)
Therefore, (2x - y)g of acidified K2SO4 is equal to:
(2x - y)g / 174 g/mol = (1000 cm^3 * 4 mol/dm^3)
Simplifying, we get:
(2x - y)g = (1000 cm^3 * 4 mol/dm^3) * 174 g/mol
(2x - y)g = 4000 * 174 / 1000
(2x - y)g = 696 g
Now, we can solve for (x + y)g by adding equations (1) and (2):
(x - y)g + (2x - y)g = 342 g + 696 g
3x - 2y = 1038 g
Since we are trying to find the value of (x + y)g, we can multiply equation (1) by 2 and subtract equation (2) from it:
2(x - y)g - (2x - y)g = 2 * 342 g - 696 g
2x - 2y - 2x + y = 684 g - 696 g
-2y + y = -12 g
y = 12 g
Now we substitute the value of y in equation (1):
(x - 12)g = 342 g / 2
(x - 12)g = 171 g
x - 12 = 171 g / g
x - 12 = 171 g
Finally, we can find the value of (x + y)g by substituting the value of y back into equation (1):
(x + 12)g = 171 g + 12 g
x + 12 = 183 g
x = 183 g - 12 g
x = 171 g
Therefore, the value of (x + y)g is 171 g + 12 g, which equals 183 g.