The density of a 2.45 M aqueous methanol solution is .976 g/ml. what is the mass of water present in 1.00 liter of this solution
density = 0.976 g/mL; therefore, 1.00 L has a mass of 0.976 g/mL x 1000 mL = 976 grams for 1L soln.
mass CH3OH = 2.45M x molar mass CH3OH = about 80 g but you need to do it more carefully.
Then mass water must be 976-80 = ?
To find the mass of water present in 1.00 liter of the given solution, we need to calculate the mass of the solution and then subtract the mass of methanol to determine the mass of water.
First, we can calculate the mass of the solution using its density. The density is given as 0.976 g/ml, so to find the mass of the solution in grams, we can multiply the density by the volume:
Mass of solution = Density × Volume
The volume is given as 1.00 liter, which is equivalent to 1000 ml. Substituting the values:
Mass of solution = 0.976 g/ml × 1000 ml
= 976 g
Next, we need to find the mass of methanol in the solution. The concentration of methanol is given as 2.45 M. Molarity is defined as moles of solute per liter of solution. Therefore, the concentration gives us the number of moles of methanol in 1.00 liter of the solution.
To find the mass of methanol, we can multiply the number of moles by the molar mass of methanol. The molar mass of methanol (CH3OH) is 32.04 g/mol.
Mass of methanol = Moles of methanol × Molar mass of methanol
Moles of methanol = Concentration × Volume
The volume is given as 1.00 liter, so:
Moles of methanol = 2.45 M × 1.00 L
= 2.45 moles
Now, we can calculate the mass of methanol:
Mass of methanol = 2.45 moles × 32.04 g/mol
= 78.438 g
Finally, we can find the mass of water by subtracting the mass of methanol from the mass of the solution:
Mass of water = Mass of solution - Mass of methanol
= 976 g - 78.438 g
= 897.562 g
Therefore, the mass of water present in 1.00 liter of the given solution is approximately 897.562 grams.