What is the density (g/mL) of each of the following samples?

b. ...

Question

What is the density (g/mL) of each of the following samples?

b. A syrup is added to an empty container with a mass of 115.25 g. When 0.100 pt of syrup is added, the total mass of the container and syrup is 182.48 g.

Accepted Answer

Welcome back, everyone determine the density in grams per milliliter of a sample of honey. In the following scenario, an empty beaker has a mass of 104.256 g. The mass of the beaker increased to 303.130 g. 1.25 pints of honey was added to the beaker. So we are given a visualization of our two beakers. Our first speaker is empty with a mass of 104.256g itself. And then our second beer beaker, sorry, Which appears almost a little more than halfway filled has the mass of 303.130g. And we can interpret that this second beaker likely contains our honey that was added inside. And so let's first recall that in order to calculate density, we would recall the symbol row which is equal to mass divided by volume of our substance. Now, we need to begin by finding our mass of honey and we can utilize the difference between our empty beaker, which is our first speaker versus our speaker with honey. And so to find the mass of honey, we're gonna take our mass of the beer plus the honey subtracted from our mass of our empty beaker. And so, because we know our first variable is the mass of 303.130 g. Our mass of beaker plus honey, we're gonna subtract this and take the mass of the empty beaker, which we're given as 104.256 g. And from this difference, we'll find that our mass of honey is 198.874 g. Next, we're going to need our volume of honey. We're told that we have 0.25 pints of honey and the beaker. And let's recall the conversion factors to go to get our volume of honey. We're going to take, I'm sorry, this should be an equal sign. We're going to take our 0.25 pints and multiply by our first conversion factor to go from pints in the denominator to quartz in the numerator. Recall that one .6, pints is equivalent to one quart in the numerator. And so canceling out pints, we're left with quarts. And again, we need our final unit of volume to be in milliliters. And so we're going to multiply by our next conversion factor to go from quartz in the denominator to leaders in the numerator. And let's recall that for 1.57 quarts. We have one liter canceling out quarts. We're going to multiply by our next conversion factor to go from leaders in the denominator to milliliter in the numerator. And let's recall that our prefix Milly tells us that we have 10 to the negative third power of our base unit leader in the denominator canceling out leaders. We're left with Millis in the numerator as our unit of volume. And we'll find that our volume of honey is equal to 142. mL. Now, with this volume of honey, let's get into calculating our density of honey row. And so we would set this equal to our massive honey which we found above as 198.874 g. This is divided by our volume of honey which we just found in milliliters as 142.53 mL. So we have units of grams per mil liter and we'll find our result as our density of honey equal to exactly 1.4 g per milliliter. And this is going to be our final answer as our density of the honey. And this will correspond to choice c in the multiple choice 1.4 g per milli. I hope this made sense. And let us know if you have any questions

What is the density (g/mL) of each of the following samples? b. A syrup is added to an empty container with a mass of 115.25 g. When 0.100 pt of syrup is added, the total mass of the container and syrup is 182.48 g.

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