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?

Answer 1

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.