Nickel carbonyl, \(\mathrm{Ni}(\mathrm{CO})_{4},\) is one of the most toxic substances known. The present maximum allowable concentration in laboratory air during an 8 -hr workday is \(1 \mathrm{ppb}\) (parts per billion) by volume, which means that there is one mole of \(\mathrm{Ni}(\mathrm{CO})_{4}\) for every \(10^{9}\) moles of gas. Assume \(24^{\circ} \mathrm{C}\) and 101.3 kPa pressure. What mass of \(\mathrm{Ni}(\mathrm{CO})_{4}\) is allowable in a laboratory room that is \(3.5 \mathrm{~m} \times 6.0 \mathrm{~m} \times 2.5 \mathrm{~m} ?\)

Short Answer

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The allowable mass of Nickel carbonyl, \(\mathrm{Ni}(\mathrm{CO})_{4}\), in a laboratory room with dimensions \(3.5 \mathrm{~m} \times 6.0 \mathrm{~m} \times 2.5 \mathrm{~m}\) at a temperature of \(24^{\circ} \mathrm{C}\) and pressure of 101.3 kPa is calculated using the following steps: 1. Determine the room's volume (V): \(V = 3.5 \times 6.0 \times 2.5 \mathrm{m}^3\) 2. Convert the volume to liters: \(V(L) = V(\mathrm{m}^3) \times 1000\) 3. Calculate the moles of air (n) in the room using the ideal gas law: \(n = PV / RT\) 4. Determine the moles of \(\mathrm{Ni}(\mathrm{CO})_{4}\) allowed in the room: Moles of \(\mathrm{Ni}(\mathrm{CO})_{4}\) \(= \frac{\text{Total moles of gas}}{10^9}\) 5. Convert the moles of \(\mathrm{Ni}(\mathrm{CO})_{4}\) to mass: Mass of \(\mathrm{Ni}(\mathrm{CO})_{4}\) \(= \text{Moles of} \; \mathrm{Ni}(\mathrm{CO})_{4} \times \text{Molar mass of} \; \mathrm{Ni}(\mathrm{CO})_{4}\) The final result will be the allowable mass of \(\mathrm{Ni}(\mathrm{CO})_{4}\) in the laboratory room.

Step by step solution

01

Calculate the total volume of the room.

The given dimensions of the room are: Length = 3.5 m Width = 6.0 m Height = 2.5 m The total volume of the room (in m³) can be calculated by multiplying the three dimensions: Volume = Length × Width × Height
02

Convert the volume to liters.

To convert the volume from m³ to liters, we will multiply by 1000: Volume (L) = Volume (m³) × 1000
03

Determine the number of moles of air using the ideal gas law.

The ideal gas law equation is given as: PV = nRT We need to solve for n (moles of air) using the given conditions: P = 101.3 kPa V = Volume (L) (from Step 2) R = 8.314 L・kPa/(K・mol) (gas constant) T = 24 °C + 273.15 = 297.15 K Rearranging the equation to solve for n: n = PV / RT
04

Calculate the moles of \(\mathrm{Ni}(\mathrm{CO})_{4}\) allowed in the room.

One ppb concentration means that there is 1 mole of \(\mathrm{Ni}(\mathrm{CO})_{4}\) for every \(10^9\) moles of gas. To find the moles of \(\mathrm{Ni}(\mathrm{CO})_{4}\) allowed in the room, we will use: Moles of \(\mathrm{Ni}(\mathrm{CO})_{4}\) = Total moles of gas / \(10^9\)
05

Convert moles of \(\mathrm{Ni}(\mathrm{CO})_{4}\) to mass.

To calculate the mass of \(\mathrm{Ni}(\mathrm{CO})_{4}\), we will multiply the moles of \(\mathrm{Ni}(\mathrm{CO})_{4}\) (from Step 4) by the molar mass of \(\mathrm{Ni}(\mathrm{CO})_{4}\). Molar mass of \(\mathrm{Ni}(\mathrm{CO})_{4}\) = (58.69 + 4(12.01 + 16.00)) g/mol Mass of \(\mathrm{Ni}(\mathrm{CO})_{4}\) = Moles of \(\mathrm{Ni}(\mathrm{CO})_{4}\) × Molar mass of \(\mathrm{Ni}(\mathrm{CO})_{4}\) The result will be the mass of \(\mathrm{Ni}(\mathrm{CO})_{4}\) allowable in the laboratory room.

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