In our lab, we heated a beaker between 60-70 degrees, adding a small test tube w

Question

In our lab, we heated a beaker between 60-70 degrees, adding a small test tube with 5.XXX g of Lauric Acid until it melted, then took it out and measured the temperature every 20 seconds until we plateaued. We did the same for 1.XXX g of Benzoic acid (added to the same test tube the Lauric Acid was added into). The starred value on trial 1 of lauric/benzoic acid is when crystals started appearing (freezing point). I have posted all exact data points, the graphs I created with seperate trendlines aiming to find the intersection (red circle). The blue sticky notes have the points of intersection for each trail. I entered each trendline equation in Wolfram Intersection points of two curves/lines calculator for the x-coordinate. I substituted the x-intercept into one of the equations to find the y, coming up with the exact point. Can someone please show me how to use these points for the following questions? I'm so confused where to go from here. Thank you so much in advance. E,periment 1: Determination of Molar Mass using Freezing Port Depression Data Sheets -Trial #1 Lauric Acid Mass of lauric acid: 5.o24 4 Time (sec) Temperature Time (sec) Temperature 400 Sec 420 13 440 538 460 480 120 50.9% 500 1400 520 160 540 180 560 200 580 220 240 620 260 640 280 660 300 680 320 45.0c 700 340 720 360 740

Explanation / Answer

1. Depression in Freezing point = (Freezing point of pure solvent) - (Freezing point of solution)

From graph,

Average freezing point of pure solvent = 44.322 and Average freezing point of solution = 38.084

therefore

Depression in freezing point = 44.322 - 38.084 = 6.238

2. molality = depression in freezing point / cryoscopic constsnt

m = 6.238 / 3.9 = 1.5994

3. moles of benzoic acid = m * mass of solvent

= 1.5994 * 0.005017 = 0.008

4. Molar mass of benzoic acid = (0.0045) / (0.008 * 0.005017) = 112.11

5. Percent error of molar mass = [(122.12 - 112.11) / 122.12 ] * 100 = 8.1886

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