Question

1. while doing the determination of the density of the liquid, sam determines the volume of the flask using distilled water at 25⁰c. john borrows the flask and sam’s calculated volume of the flask and uses it to determine the density of his unknown liquid at 20⁰c. what, if any, specific effect will this have on the results of john’s density determination? explain your answer.

Explanation:

1. Recall the relationship between temperature, volume, and density. Substances (including the flask material) expand with increasing temperature and contract with decreasing temperature.
2. Sam determined the flask's volume at \(25^\circ \text{C}\). The flask is made of a material (e.g., glass) that expands when heated and contracts when cooled. At \(25^\circ \text{C}\), the flask has a certain volume \(V_{25}\).
3. John uses the flask at \(20^\circ \text{C}\), which is lower than \(25^\circ \text{C}\). Due to thermal contraction, the actual volume of the flask at \(20^\circ \text{C}\) (\(V_{20}\)) is less than \(V_{25}\) (the volume Sam calculated).
4. The formula for density is \(

ho=\frac{m}{V}\), where \(m\) is mass and \(V\) is volume. John measures the mass of his unknown liquid using the flask. Let's assume the mass of the liquid is \(m\) (which is correct for the actual volume at \(20^\circ \text{C}\)). But he uses the volume \(V_{25}\) (from Sam's calculation) instead of the actual volume \(V_{20}\) (which is smaller) in his density calculation.

1. Since \(V_{25}>V_{20}\) and \(

ho = \frac{m}{V}\), when \(V\) is overestimated (using \(V_{25}\) instead of \(V_{20}\)), the calculated density \(
ho_{\text{calculated}}=\frac{m}{V_{25}}\) will be less than the actual density \(
ho_{\text{actual}}=\frac{m}{V_{20}}\) (because dividing by a larger number gives a smaller result).

Answer:

John's calculated density of the unknown liquid will be lower than the actual density. This is because the flask contracts at \(20^\circ \text{C}\) (lower than \(25^\circ \text{C}\)) so its actual volume at \(20^\circ \text{C}\) is smaller than the volume Sam calculated at \(25^\circ \text{C}\). Using the larger (incorrect) volume from Sam in the density formula \(
ho=\frac{m}{V}\) (where \(m\) is the mass of the liquid) leads to a lower calculated density, as density is inversely proportional to the volume used in the calculation when mass is constant.