a) A quantidade de mols será:

\(\begin{align} & \frac{19}{14,3}=\frac{1}{x} \\ & x=0,75mol \\ \end{align}\ \)

b)  A quantidade de molécula de testosterona será:

\(\begin{align} & \frac{28}{14,3}=\frac{1}{x} \\ & x=0,5mol \\ \end{align}\ \)

c) \(6,02\cdot {{10}^{23}}\) mol

d)

A massa será:

\(\begin{align} & \frac{28}{x}=\frac{6,02\cdot {{10}^{23}}}{3,02\cdot {{10}^{23}}} \\ & x=(3,02\cdot {{10}^{23}}\cdot 28)/6,02\cdot {{10}^{23}} \\ & x=14,3g \\ \end{align}\ \)