Yeah. So the pipettes that you'll be using in your analytical labs come from the manufacturer as usually class a pipettes. But once you obtain them there is some level of uncertainty associated with them, you have to first calibrate them before we can use them in experiments. So here we're gonna say convenient method and calibrating pipettes is to weigh the water delivered from them by using the density of water at a given temperature. We can determine the volume of density with greater accuracy. So here it says, assume a 50 millimeter pipette is in need millim calibration. An empty flask weighs 49.563g. When water delivered from the pipette is added to the empty flask, the new mass is recorded as 69.618g were asked what is the mass of water delivered? Now here they're saying the density of the standard weights is determined to be 8.40 g per mil leader. Remember in the equation above, we said that the standard weight is 8.0 g per mil leader in this question, we must be using a different type of metal or alloy for those calibration weights, which is why it's a new number. So again, always be careful if they're giving you a new density for the calibration weights, if they do use that one. So we're gonna say here em equals and prime times one minus D. A over D. W Divide by 1 - D. A over D. We need to first find the apparent mass. Well, when the flask is filled with water, The weight of the water and the flask together is 69.618g. When the flask is empty, it weighs that much. When we subtract them, That'll give me the mass of just the water. So that comes out to 20.055 g. So that's our apparent mass. So times one over. So d. is the density of air which is .0012 g per meal leader. The density of the calibration weight we're g is 8.40 g per male leader, Divided by 1 -4 again divided by the density of water. Here, they don't tell us the exact temperature. The density of water can change with changing temperatures but most of them are around one g per mil leader. So we're gonna use one grand per millimeter for the density of water. So we just have to solve for true mass now. So it's 20.055g. Times when I work all this out in here, It gives me a .999857, divide by the bottom. When I subtract this value here from one, what does it give me? Give me 10.9988. Then what do we get at the end? So at the end we're gonna get 20.079 g. So notice are apparent mass. When we read it from the analytical balance was that number. But the true mass. when we take into account air flow gives us this new mass for our water. So again, this is a great way in order to find the true mass of an object, and also a good way to calibrate the pipettes that you'll be using extensively within your analytical labs. So now that we've seen this will take a look at buoyancy when it's found within fluids.
