Question

A solution is prepared by adding 91.9 g of ethanol, ch3ch2oh, to 481.0 ml of water. the final volume of the solution is 597.5 ml. assuming the density of water is 1.0 g/ml and the density of ethanol is 0.789 g/ml, calculate the molality of the ethanol in the solution.

Answer 1

17.51M

Step-by-step explanation:

first we make out what we've been given .

Mass of ethanol= 91.9g

volume of water = 481.0ml

volume of solution= 597.5ml

density of water = 1.0g/ml

density of ethanol= 0.789g/ml.

firstly find the volume of ethanol since the total volume of the solution is 597.5ml and we've been given the volume of water which is 481.0ml , we subtract the volume of water from the volume of the solution , to find out the volume of ethanol .

volume e =597.5ml - 481.0ml

= 116.5ml

Then we find the molecular mass of ethanol , which is 45.04g/mol .

then we find the number of moles of ethanol .

n=mass /molar mass

=91.9g/ 45.04g/mol

=2.04mols .

Then convert the volume from milliliters to liters , by dividing by a 1000 .

v= 116.5/1000

=0.1165L

since we have our number of moles and volume , we can now find the molarity . which is

M=n/v

= 2.04mols /0.1165L

= 17.51M .