Answer:
[tex]\large \boxed{\text{69.3 mg}}[/tex]
Explanation:
1. Volume of sealed tube
Assume the sealed tube is a right circular cylinder in which the cap and the base are also 4.20 mm thick.
Its outside dimensions are 155 mm long × 10.0 mm diameter.
Its inside dimensions are
h = 155 mm - 2 × 4.20 mm = 146.6 mm
r = 5.0 mm - 4.20 mm = 0.8 mm
V = πr²h = π(0.8)²× 146.6 mm³ = 294.8 mm³ = 0.2948 cm³
2. Calculate the mass of NH₃
[tex]\begin{array}{rcl}\text{Density} & = & \dfrac{\text{Mass}}{\text{Volume}}\\\\\rho & = &\dfrac{m}{V}\\\\0.235 \text{ g$\cdot$ cm}^{-3} & = & \dfrac{m}{\text{0.2948 cm}^{3}}\\\\m & = & \text{0.0693 g}\\& = & \textbf{69.3 mg}\\\end{array}\\\text{You must seal $\large \boxed{\textbf{69.3 mg}}$ of ammonia in the tube.}[/tex]
