Questions 1.44 - Chemistry: Molecular Nature Of Matter And Change

Question

A 25.0-g sample of each of three unknown metals is added to 25.0 mL of water in graduated cylinders A, B, and C, and the final volumes are depicted in the circles below. Given their densities, identify the metal in each cylinder: zinc ( 7.14g/ml ), iron ( 7.87g/ml ), nickel ( 8.91g/ml ).

Step-by-Step Solution

Answer

Answer

The metal present in cylinder A is Iron, B is Nickel, and C is zinc.

Definition

The formula for the density of metal in the cylinder is: 

$\mathrm{Density} = \frac{\mathrm{Mass} \mathrm{of} \mathrm{the} \mathrm{metal}}{\mathrm{Volume} \mathrm{of} \mathrm{the} \mathrm{metal}}$

  

There is a rise in water volume once you add metal. The difference in volume gives metal volume added.

Cylinder A

The volume of water in cylinders A, B, & C is 25 mL. Water’s volume inside cylinder after metal addition is 28.2 mL in A, 27.9 mL in B, and 28.5 mL in C.

The volume of metal can be denoted as:

$\begin{array}{rcl}\mathrm{Volume} \mathrm{of} \mathrm{Metal} & =& {\mathrm{V}}_{\mathrm{after}} - {\mathrm{V}}_{\mathrm{before}}\\ & =& \left(28.2-25.0\right)\mathrm{ml}\\ & =& 3.2\mathrm{ml}\\ & & \\ & & \mathrm{The} \mathrm{density} \mathrm{of} \mathrm{the} \mathrm{metal} \mathrm{can} \mathrm{be} \mathrm{calculated} \mathrm{as}:\\ \mathrm{Density} & =& \frac{\mathrm{mass} \mathrm{of} \mathrm{metal}}{\mathrm{volume} \mathrm{of} \mathrm{metal}}\\ & =& \frac{25\mathrm{g}}{3.2\mathrm{ml}}\\ & & 7.81\mathrm{g}/\mathrm{ml}\end{array}$

The density of the metal calculated above is equal to the iron’s density. Thus, the metal in cylinder A is iron.

Cylinder B

$\begin{array}{rcl}& & \mathrm{The} \mathrm{volume} \mathrm{of} \mathrm{metal} \mathrm{can} \mathrm{be} \mathrm{denoted} \mathrm{as}:\\ \mathrm{Volume} \mathrm{of} \mathrm{metal} & =& {\mathrm{V}}_{\mathrm{after}} - {\mathrm{V}}_{\mathrm{before}}\\ & =& \left(2.79 - 25.0\right)\mathrm{ml}\\ & =& 2.9\mathrm{ml}\\ & & \\ & & \mathrm{The} \mathrm{density} \mathrm{of} \mathrm{the} \mathrm{metal} \mathrm{can} \mathrm{be} \mathrm{calculated} \mathrm{as}:\\ \mathrm{Density} & =& \frac{\mathrm{Mass} \mathrm{of} \mathrm{metal}}{\mathrm{Volume} \mathrm{of} \mathrm{metal}}\\ & =& \frac{25\mathrm{g}}{2.9\mathrm{ml}}\\ & =& 8.62\mathrm{g}/\mathrm{ml}\end{array}$

The density of the metal calculated above is approximately equal to the iron’s density (8.91 g/mL). Thus, the metal in cylinder B is Nickel.

Cylinder C

$\begin{array}{rcl}& & \mathrm{The} \mathrm{volume} \mathrm{of} \mathrm{metal} \mathrm{can} \mathrm{be} \mathrm{denoted} \mathrm{as}: \\ \mathrm{Volume} \mathrm{of} \mathrm{metal} & =& {\mathrm{V}}_{\mathrm{after}} - {\mathrm{V}}_{\mathrm{before}}\\ & =& \left(28.5 - 25.0\right)\mathrm{ml}\\ & =& 3.5\mathrm{ml}\\ & & \\ & & \mathrm{The} \mathrm{density} \mathrm{of} \mathrm{the} \mathrm{metal} \mathrm{can} \mathrm{be} \mathrm{calculated} \mathrm{as}:\\ \mathrm{Density} & =& \frac{\mathrm{Mass} \mathrm{of} \mathrm{metal}}{\mathrm{Volume} \mathrm{of} \mathrm{metal}}\\ & =& \frac{25\mathrm{g}}{3.5\mathrm{ml}}\\ & =& 7.14\mathrm{g}/\mathrm{ml}\end{array}$

The density of the metal calculated above is equal to the iron’s density. Thus, the metal in cylinder C is Zinc.