# A graduated cylinder contains 225 mL of water. What is the new water level affer 30.2 g of silver metal with a density of 10.5 g/mL is submerged in the water?

Jan 21, 2017

Well, we work out the water displaced by the silver........

#### Explanation:

Well, we work out the water displaced by the silver........by calculating the volume of the silver nugget using the density.

$\text{Density} \left(\rho\right)$ $=$ $\text{Mass"/"Volume}$

Thus $\frac{\text{Volume"="Mass}}{\rho}$

$=$ $\frac{30.2 \cdot g}{10.5 \cdot g \cdot m {L}^{-} 1}$ $\cong$ $3 \cdot m L$.

This is consistent dimensionally, because $\frac{\cancel{g}}{\cancel{g} \cdot m {L}^{-} 1} = \frac{1}{\frac{1}{m {L}^{-} 1}} = m L$ as required.

So the volume will be approx. $228 \cdot m L$