Questions: INTRODUCTION TO CHEMISTRY (CHM090) Prof. Raid Albattah Experiment 5: Determination of Density Lab - Part A Substance (water) - Part B Substance (Unknown Liquid) - Part C Substance (solid object) Mass of the Beaker: 20 mL (Part A), 10 mL (Part B), lead (Part C) Mass of the beaker + substance: 31.34 mL (Part A), 41.34 mL (Part B), 11.340 (Part C) Mass of the substance: 20 mL (Part A), 10 mL (Part B), 11.340 (Part C) Volume of the substance: Density of the substance (d=M / V):

Solution Steps

To find the mass of each substance, subtract the mass of the beaker from the mass of the beaker plus the substance.

• Part A (Water): \[ \text{Mass of water} = 31.34 \, \text{g} - 20 \, \text{g} = 11.34 \, \text{g} \]

• Part B (Unknown Liquid): \[ \text{Mass of unknown liquid} = 41.34 \, \text{g} - 10 \, \text{g} = 31.34 \, \text{g} \]

• Part C (Solid Object - Lead): The mass of the solid object is already given as 11.340 g.

For Part A and Part B, the volume is given directly as the volume of the liquid. For Part C, we need to calculate the volume using the mass and density of lead.

• Part A (Water): \[ \text{Volume of water} = 20 \, \text{mL} \]

• Part B (Unknown Liquid): \[ \text{Volume of unknown liquid} = 10 \, \text{mL} \]

• Part C (Solid Object - Lead): The density of lead is approximately \(11.34 \, \text{g/mL}\). Using the formula \(d = \frac{M}{V}\), we can find the volume: \[ V = \frac{M}{d} = \frac{11.340 \, \text{g}}{11.34 \, \text{g/mL}} = 1.000 \, \text{mL} \]

Use the formula \(d = \frac{M}{V}\) to calculate the density.

• Part A (Water): \[ d = \frac{11.34 \, \text{g}}{20 \, \text{mL}} = 0.5670 \, \text{g/mL} \]

• Part B (Unknown Liquid): \[ d = \frac{31.34 \, \text{g}}{10 \, \text{mL}} = 3.134 \, \text{g/mL} \]

• Part C (Solid Object - Lead): \[ d = \frac{11.340 \, \text{g}}{1.000 \, \text{mL}} = 11.34 \, \text{g/mL} \]

Final Answer