An solution of antifreeze is prepared by mixing 23.0mL of ethylene glycol (d = 1.11 g/mL;molar mass = 62.07 g/mol) with 50.0 mL H2O (d = 1.00 g/mL) at 25°C. If the density of the antifreeze solution is 1.07 g/mL, what is its molaritY?
Well, let's calculate it together! To find the molarity, we need to determine the number of moles of ethylene glycol (C2H6O2) in the solution.
First, let's find the mass of ethylene glycol (C2H6O2):
Mass of ethylene glycol = volume x density
Mass of ethylene glycol = 23.0 mL x 1.11 g/mL
Mass of ethylene glycol = 25.53g
Next, let's find the number of moles of ethylene glycol:
Number of moles = mass / molar mass
Number of moles = 25.53g / 62.07 g/mol
Number of moles = 0.411 mol
Now, let's calculate the total volume of the solution:
Total volume = volume of ethylene glycol + volume of water
Total volume = 23.0 mL + 50.0 mL
Total volume = 73.0 mL
Finally, we can calculate the molarity (M) using the formula:
Molarity = number of moles / volume of solution in liters
Molarity = 0.411 mol / (73.0 mL/1000)
Molarity = 5.64 M
So, the molarity of the antifreeze solution is 5.64 M. Keep in mind that being such a highly concentrated solution, you might want to avoid drinking it and sticking to some cool jokes instead!
To find the molarity of the antifreeze solution, we need to first determine the number of moles of ethylene glycol. Here are the steps to calculate it:
Step 1: Calculate the mass of ethylene glycol:
Mass of ethylene glycol = Volume of ethylene glycol x Density of ethylene glycol
Mass of ethylene glycol = 23.0 mL x 1.11 g/mL
Mass of ethylene glycol = 25.53 g (rounded to 2 decimal places)
Step 2: Calculate the number of moles of ethylene glycol:
Number of moles of ethylene glycol = Mass of ethylene glycol / Molar mass of ethylene glycol
Number of moles of ethylene glycol = 25.53 g / 62.07 g/mol
Number of moles of ethylene glycol = 0.4116 mol (rounded to 4 decimal places)
Now, let's find the volume of the antifreeze solution:
Step 3: Calculate the total volume of the antifreeze solution:
Total volume of antifreeze solution = Volume of ethylene glycol + Volume of water
Total volume of antifreeze solution = 23.0 mL + 50.0 mL
Total volume of antifreeze solution = 73.0 mL (rounded to 1 decimal place)
Finally, let's calculate the molarity:
Step 4: Calculate the molarity of the antifreeze solution:
Molarity = Number of moles of solute / Volume of solution (in liters)
Molarity = 0.4116 mol / (73.0 mL x (1 L / 1000 mL))
Molarity = 0.00564 M (rounded to 3 decimal places)
The molarity of the antifreeze solution is 0.00564 M.
To find the molarity of the antifreeze solution, we need to know the number of moles of the solute (ethylene glycol) and the volume of the solution.
First, let's find the number of moles of ethylene glycol.
1. Calculate the mass of ethylene glycol:
Mass of ethylene glycol = volume of ethylene glycol x density of ethylene glycol
= 23.0 mL x 1.11 g/mL
= 25.53 g
2. Calculate the number of moles of ethylene glycol using its molar mass:
Moles of ethylene glycol = mass of ethylene glycol / molar mass of ethylene glycol
= 25.53 g / 62.07 g/mol
= 0.4117 mol
Next, let's find the volume of the antifreeze solution.
3. Calculate the total mass of the antifreeze solution:
Mass of antifreeze solution = (mass of ethylene glycol + mass of water)
= (25.53 g + volume of water x density of water)
4. Rearrange the equation to solve for the volume of water:
volume of water = (mass of antifreeze solution - mass of ethylene glycol) / density of water
= (50.0 mL x 1.00 g/mL - 25.53 g) / 1.00 g/mL
= 24.47 mL
5. Calculate the volume of the antifreeze solution by adding the volumes of ethylene glycol and water:
volume of antifreeze solution = volume of ethylene glycol + volume of water
= 23.0 mL + 24.47 mL
= 47.47 mL
Finally, let's calculate the molarity of the antifreeze solution.
6. Convert the volume of the antifreeze solution to liters:
volume of antifreeze solution = 47.47 mL * (1 L / 1000 mL)
= 0.04747 L
7. Calculate the molarity of the antifreeze solution using the number of moles and volume:
Molarity = moles of ethylene glycol / volume of antifreeze solution
= 0.4117 mol / 0.04747 L
= 8.67 M
Therefore, the molarity of the antifreeze solution is approximately 8.67 M.
Tricky...I don't want to write out ethylene glycol all the time so let's let eth stand for that.
M solution = mols eth/L solution.
How to get values to plug in.
mols eth = gram/molar mass and you get grams from mass eth = volume eth x density eth. You know volume and density; solve for mass, then for mols use mols eth = grams eth/molar mass eth.
Now for the L solution. The tricky part is that SOME students (of course you wouldn't do that) will add 23 mL to 50 mL H2O to get final volume of 73 mL BUT that isn't right because volumes are not additive. What you do is this. You know grams eth from that 23.0 mL when you did the volume to mass conversion and you know grams of the 50.00 mL (from that density calculation)(50 mL H2O at density of 1.00 g/mL will weigh 50 grams), and you know the density of the FINAL solution so calculate final volume from that. Now you have mols eth in the final solution and the volume of the final solution. You can calculate the M of the final solution.