A 500ml saturated solution of MgCO3 (M=84) is reduced to 120 ml by evaporation. What mass of solid MgCO3 is formed?
The answer is 0.20g, I just need a detailed explanation.
Thanks!
OK. Thanks.
MgCO3 ==> Mg^+2 + CO3^-2
..x........x.......x
Ksp = (Mg^+2)(CO3^-2) - (x)(x)
x = sqrt(4E-5) = 0.00632 moles/L or 0.00632 x 84 = 0.531 g/L
In 500 mL there will be 1/2 that or .265 grams.
If we evaporate water until the volume is 120 mL, then 0.531 g/L becomes
0.531 g/L x (120 mL/1000 mL) = 0.0637 g.
We started with 0.265 g in 500
............less 0.0637 in 120 mL =
.................-------
solid ppting.....0.201 g which rounds to 0.2 g to 1 significant figures (from the Ksp value).
Do you have a solubility or a Ksp to use. There is a good bit of disagreement on the net about the correct value for Ksp for MgCO3.
Oh sorry the Ksp=4x10^(-5)
Thanks! But im confused on why you would subtract the .0637g in 120mL from .265g in 500mL. Why don't you do (grams from 500mL-grams from 380mL)?
Well, let's break this down step by step.
First, we need to find out the initial mass of MgCO3 present in the 500 ml saturated solution. To do this, we can use the formula:
Mass = (Concentration * Volume) / Molar Mass
The concentration of a saturated solution is the maximum amount of solute that can dissolve in a given amount of solvent at a given temperature. In this case, the concentration of MgCO3 is at its maximum, so we can use its solubility. The solubility of MgCO3 is 0.067 g/ml.
So, the initial mass of MgCO3 in the 500 ml solution is:
Mass = (0.067 g/ml * 500 ml) / 84 g/mol
Mass = 0.399 g
Now, we know that the solution was reduced to 120 ml by evaporation. We need to find the mass of the solid formed during this process. To do this, we can calculate the mass of the remaining solution and subtract it from the initial mass.
The final mass of the remaining solution is:
Final Mass = (0.067 g/ml * 120 ml) / 84 g/mol
Final Mass = 0.096 g
Now, to find the mass of solid MgCO3 formed, we subtract the final mass of the solution from the initial mass:
Mass of Solid = Initial Mass - Final Mass
Mass of Solid = 0.399 g - 0.096 g
Mass of Solid = 0.303 g
So, the mass of solid MgCO3 formed is 0.303 g. I'm sorry if this explanation was a bit dry, but I hope you found it useful!
To solve this problem, we need to use the concept of molarity and the mole-to-mass relationship.
1. Calculate the initial moles of MgCO3 in the saturated solution.
- The molar mass of MgCO3 is 84 g/mol.
- Knowing that the solution is saturated, we can assume that all of the solute (MgCO3) is dissolved in the solution.
- The molarity (M) of a solution is defined as moles of solute per liter of solution.
- Given that the volume of the solution is 500 ml, which is equal to 0.5 L, and that it is saturated, the molarity can be calculated as follows:
Molarity (M) = moles of solute / volume of solution in liters
M = moles of MgCO3 / 0.5 L
- From the given information, we can rearrange the equation to solve for the moles of MgCO3:
Moles of MgCO3 = M x volume of solution in liters
Moles of MgCO3 = M x 0.5 L
- Now, substitute the given values:
Moles of MgCO3 = M x 0.5 L = M x 0.5
2. Calculate the moles of MgCO3 remaining after evaporation.
- The solution volume is reduced to 120 ml, which is equal to 0.12 L.
- Using the same equation as above, we can calculate the moles remaining in the solution:
Moles of MgCO3 remaining = M x volume of solution after evaporation in liters
- Substitute the given values:
Moles of MgCO3 remaining = M x 0.12 L
3. Calculate the mass of solid MgCO3 formed.
- Since the solid MgCO3 is no longer dissolved in the solution after evaporation, the moles of MgCO3 remaining are the same as the moles of solid MgCO3 formed.
- The relationship between moles and mass is given by the molar mass (M) of the compound.
- The mass of solid MgCO3 formed can be calculated as:
Mass = Moles x Molar mass
- Substitute the given values:
Mass = Moles of MgCO3 remaining x Molar mass of MgCO3
Mass = Moles of MgCO3 remaining x 84 g/mol
- Now, substitute the moles of MgCO3 remaining from step 2 into the equation:
Mass = (M x 0.12) x 84 g/mol
Mass = M x 0.12 x 84 g/mol
- Finally, substitute the value of M into the equation:
Mass = 0.2 x 0.12 x 84 g
- Calculate the mass:
Mass = 0.20 g
Therefore, the mass of solid MgCO3 formed after 120 ml of evaporation is 0.20 grams.
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(10 points possible)
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/* 1-5 */
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[Ar]4s23d104p5
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