In: Chemistry
1. A student prepared a solution of ethylene glycol in distilled water. He mixed the 5ml of ethylene glycol with 45 ml water. The mass of the ethylene glycol was 5.580g. The total mass of the solution was 49.037 g.
a. Calculate the mass percent of ethylene glycol in the solution.
the student determined the average mass of 10.00 ml of this solution is 10.063g
b. calculate the average density of the solution
c. in a similar fashion, the student prepared other aqueous solutions of ethylene glycol and measured thier densities, obtaining the data shown in the following.
mass percent ethylene glycol, % 0.00 (answer from part A goes here) 22.75 33.22
average density, g/ml 0.993 (answer from part B goes here) 1.021 1.030
d. Plot average density versus mass percent ethylene glycol, and use the best straight line through the data points.
e. An aqueous solution of ethylene glycol has a density of 1.010g/ml What is the mass percent ethylene glycol in this solution?
PLEASE SHOW WORK FOR CALCULATIONS SO I MAY SEE HOW THE ANSWET IS ACHIEVED PLEASE
2. A student perfroming this experiment for solutions of sugar in water plotted her data as described in the Procedure. She found that the slope of her trend line was 0.00401g/ml *% and the y intercept was 0.995 g/ml. She also found the average density of a lemonade sample to be 1.031 g/ml. Calculate mass percent sugar in this lemonade sample.
AGAIN PLEASE SHOW WORK FOR CALCULATIONS SO I MAY SEE HOW THE ANSWET IS ACHIEVED PLEASE
THANK YOU
1. ethylene glycol (5.580 g) in distilled water.
a. mass % of ethylene glycol = (mass of ethylene glycol/total mass of solution) x 100
= (5.580/49.037) x 100
= 11.38 %
b. Average density of solution = average mass/volume of solution
= 10.063/10
= 1.0063 g/ml
c. The data collected were,
mass (%) Ethylene glycol : 0.00 11.38 22.75 33.22
average density (g/ml) : 0.993 1.0063 1.021 1.030
d. A plot of average density versus mass % is shown below.
e. Mass % ethylene glycol for average density 1.010 g/ml = 1.010 x 22.75/1.021 = 22.50 g/ml
2. with slope = 0.00401 g/ml^% and y-intercept 0.995 g/ml
the equation becomes,
y = 0.00401x + 0.995
with average density = 1.031, mass % sugar in solution = 1.031 = 0.00401x + 0.995
x = 8.98 %