Question

1.3 chemical problem solving
the mass of a graduated cylinder containing 21.0 ml of water is 29.1 g. an unknown metal was placed in the cylinder and completely submerged in the water. the volume of water rose to 28.6 ml. the net weight of the water, the graduated cylinder, and the unknown metal was determined to be 46.4 g. what is the density of the metal?
2.8 g/ml
3.8 g/ml
2.3 g/ml
6.1 g/ml
1.6 g/ml

Explanation:

Step1: Find the mass of the metal

The mass of the cylinder and water is 29.1 g, and the mass of the cylinder, water, and metal is 46.4 g. So the mass of the metal ($m$) is $46.4 - 29.1 = 17.3$ g.

Step2: Find the volume of the metal

The initial volume of water is 21.0 mL, and the final volume is 28.6 mL. The volume of the metal ($V$) is the change in volume, so $V = 28.6 - 21.0 = 7.6$ mL.

Step3: Calculate the density of the metal

Density ($
ho$) is given by the formula $
ho = \frac{m}{V}$. Substituting the values, we get $
ho = \frac{17.3}{7.6} \approx 2.3$ g/mL. Wait, no, wait, let's recalculate: 46.4 - 29.1 is 17.3? Wait, 46.4 - 29.1: 46.4 - 29 = 17.4, minus 0.1 is 17.3. Then 28.6 - 21.0 is 7.6. Then 17.3 divided by 7.6: 17.3 ÷ 7.6 ≈ 2.276, which is approximately 2.3? Wait, but let's check again. Wait, maybe I made a mistake. Wait, 46.4 - 29.1: 29.1 + 17.3 = 46.4, that's correct. 28.6 - 21.0 = 7.6, correct. Then 17.3 / 7.6 ≈ 2.276, which is approximately 2.3 g/mL. But wait, let's check the options. One of the options is 2.3 g/mL. Wait, but let me check again. Wait, maybe I miscalculated the mass. Wait, the net weight of water, cylinder, and metal is 46.4 g. The mass of cylinder and water is 29.1 g. So metal mass is 46.4 - 29.1 = 17.3 g. Volume of metal is 28.6 - 21.0 = 7.6 mL. Then density is 17.3 / 7.6 ≈ 2.276, which is approximately 2.3 g/mL. So the answer should be 2.3 g/mL.

Answer:

2.3 g/mL (Option: 2.3 g/mL)