water was added to 120.0g of a salt MCl2, to produce 60.0cm^3 of a saturated solution at 25celsius. its solubility at 25 celsius is 8.0moldm^-3. calculate the mass of the salt which remained undissolved
74.1
Oh, salt MCl2, you're always doing your own thing! Alright, let's crunch some numbers here.
First, we need to calculate the number of moles of the salt that dissolved in the solution. To do that, we multiply the volume of the solution (60.0 cm^3) by the solubility (8.0 mol/dm^3), but we need to convert the volume to dm^3 for consistency.
60.0 cm^3 = 0.06 dm^3
So, the number of moles of MCl2 that dissolved is:
moles = volume x solubility
moles = 0.06 dm^3 x 8.0 mol/dm^3
Nice! Now we need to convert this number of moles into grams using the molar mass of MCl2. Without the molar mass, we'd be quite lost.
However, since you didn't provide the molar mass of MCl2, I'm afraid I can't calculate the mass of the salt that remained undissolved. Could you please provide the molar mass of MCl2 so we can continue our mathemagical journey?
To calculate the mass of the salt that remained undissolved, we need to determine the amount of salt that dissolved in the water in the saturated solution and subtract it from the initial mass of the salt.
1. Determine the moles of salt dissolved in the solution:
Moles of solute = Concentration (in moldm^-3) × Volume (in dm^3)
Moles of solute = 8.0 moldm^-3 × 0.060 dm^3 (converting cm^3 to dm^3)
Moles of solute = 0.48 mol
2. Calculate the mass of the salt dissolved:
Mass of solute = Moles of solute × molar mass of salt (MCl2)
Molar mass of MCl2 = molar mass of M (atomic mass of M) + 2 × molar mass of Cl
Assuming the atomic mass of M is x and the atomic mass of Cl is 35.5 g/mol:
Molar mass of MCl2 = x + 2 × 35.5 g/mol
Since the atomic mass of M is not provided, we can't calculate the exact molar mass of the salt from the given information.
However, if we assume the salt to be MCl2, where M is a metal with the atomic mass of 24.3 g/mol (the closest value to estimate),
then the molar mass would be 24.3 g/mol + 2 × 35.5 g/mol = 95.3 g/mol.
Mass of solute(MCl2) = 0.48 mol × 95.3 g/mol = 45.8 g (approximately)
3. Finally, calculate the mass of the salt that remained undissolved:
Mass of undissolved salt = Initial mass of salt - Mass of solute
Mass of undissolved salt = 120.0 g - 45.8 g
Mass of undissolved salt = 74.2 g
Therefore, approximately 74.2 grams of the salt remained undissolved in the solution.
To calculate the mass of the salt that remained undissolved, we need to first calculate the amount of salt that dissolved in the water. Here's how you can do it:
1. Calculate the number of moles of the salt that dissolved:
To do this, we will use the equation:
Moles = Volume x Concentration
Concentration is given as 8.0 moldm^-3, and the volume is given as 60.0 cm^3. We need to convert cm^3 to dm^3, so divide the volume by 1000:
Volume = 60.0 cm^3 ÷ 1000 = 0.060 dm^3
Now, substitute the values into the equation:
Moles = 0.060 dm^3 x 8.0 moldm^-3 = 0.48 moles
2. Calculate the molar mass of the salt:
Since the salt is MCl2, we assume that "M" represents a metal ion and chlorine has a molar mass of 35.5 g/mol. Therefore, the molar mass of the salt MCl2 is the molar mass of "M" plus two times the molar mass of chlorine.
Let's assume x as the molar mass of M:
Molar mass of the salt = x + 2(35.5) = x + 71 g/mol
3. Calculate the mass of the salt that dissolved:
Using the equation:
Mass = Moles x Molar mass
Substitute the values:
Mass = 0.48 moles x (x + 71) g/mol
4. Calculate the mass of the salt that remained undissolved:
The total mass of the salt was given as 120.0 g. To find the mass of the salt that remained undissolved, subtract the mass of the salt that dissolved from the total mass of the salt:
Mass of undissolved salt = Total mass - Mass of dissolved salt
Mass of undissolved salt = 120.0 g - (0.48 moles x (x + 71) g/mol)
Note: To solve for the value of x, we would need additional information or an equation that relates the solubility of the salt at a particular temperature to the molar mass of the salt.