Solving displacement of volume in water after a object is dropped inside the water container? [closed]








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> A $\pu{53.7 g}$ object having a density of $\pu{5.39 g mL-1}$ is added to a graduated cylinder containing water and sinks to the bottom. The water level is now at the $\pu{82.2 mL}$ mark. What is the volume of the graduated cylinder?
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My work so far:


$$V = \frac{m}{\rho} = \frac{\pu{53.7 g}}{\pu{5.39 g mL-1}} = \pu{9.96 mL}$$


Should I just take the displacement of the water with the object sunk to the bottom to arrive at the original volume of water like so:


$$\Delta V = \pu{82.2 mL} - \pu{9.96 mL} = \pu{76.2 mL}$$


If not, what do I need to do to finish this problem?




Short answer: **yes**, you are correct.


You calculated the Volume of the object, which was added to the water in the cylinder to give the total volume of 82.2 mL. So deduct this and get your solution... which you did.


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