1. pure ethanol has a density of 0.78945 g/cm 3 at20.0 C and 0.80625 g/cm 3 at 0
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Question
1. pure ethanol has a density of 0.78945 g/cm3 at20.0 C and 0.80625 g/cm3 at 0.0 C. What will be thedifference in mass that 500.0 mL flask will hold at 0.0 C from thatat 20.0 C? Assume the volume of the sample at 0.0 C is exactly thesame as the volume of the sample at 20.0 C 2. A flask filled with ethanol at 20.0 C has a total mass of61.742 g. the flask is emptied and some ball bearings with a massof 21.784 g are added to the flask. The flask is topped off withethanol and now the total mass is 81.604 g. Callculate the densityof the ball bearings at 20.0 C. The density of ethanol at thistemperature is 0.78945 g/cm3. please help me step by step please. i dont understand where tostart off the problems!! 1. pure ethanol has a density of 0.78945 g/cm3 at20.0 C and 0.80625 g/cm3 at 0.0 C. What will be thedifference in mass that 500.0 mL flask will hold at 0.0 C from thatat 20.0 C? Assume the volume of the sample at 0.0 C is exactly thesame as the volume of the sample at 20.0 C 2. A flask filled with ethanol at 20.0 C has a total mass of61.742 g. the flask is emptied and some ball bearings with a massof 21.784 g are added to the flask. The flask is topped off withethanol and now the total mass is 81.604 g. Callculate the densityof the ball bearings at 20.0 C. The density of ethanol at thistemperature is 0.78945 g/cm3. please help me step by step please. i dont understand where tostart off the problems!! please help me step by step please. i dont understand where tostart off the problems!!Explanation / Answer
At 0.0 C, Vol = 500.0 mL Density = 0.80625 g/cm^3 Mass = d * v = 0.80625 g / cm^3 *500.0 mL = 403.125 g . At 20.0 C, Mass = 0.78945 g/cm^3 * 500.0 mL = 394.725 g . Difference in mass = 403.125 - 394.725 g = 8.40 g . 2 Mass of flask + ethanol = 61.742 g mass of flask + bearings + ethanol = 81.604 g Mass of bearings = 21.784 g So mass of ethanol + flask during 2nd filling = 81.604 - 21.784 g =59.82 g . Difference in mass of ethanol = 61.742 g - 59.82 g = 1.922 g . Volume of this mass of ethanol = mass / density = 1.922 g / 0.78945 g/cm^3 = 2.435 mL . This is the volume that corresponds to the volume of the ballbearings. . Hence density of ball bearings = mass / volume = 21.784 g / 2.435 mL = 8.95 g/cm^3
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