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Chapter 10: Problem 106

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 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 1.00 atm pressure. What mass of \(\mathrm{Ni}(\mathrm{CO})_{4}\) is allowable in a laboratory room that is \(12 \mathrm{ft} \times 20 \mathrm{ft} \times 9 \mathrm{ft}\) ?

Short Answer

Expert verified
The allowable mass of Ni(CO)₄ in the laboratory room under the given conditions is approximately 0.42446 grams.

Step by step solution

01

Convert room dimensions to meters

To convert the room dimensions from feet to meters, we can use the conversion factor: 1 ft = 0.3048 meters. Length = 12 ft × 0.3048 m/ft = 3.6576 m Width = 20 ft × 0.3048 m/ft = 6.0960 m Height = 9 ft × 0.3048 m/ft = 2.7432 m
02

Calculate the volume of the room

Multiply the length, width, and height to find the volume of the room in cubic meters. Volume = Length × Width × Height = 3.6576 m × 6.0960 m × 2.7432 m = 61.3084 m³
03

Find the total number of moles of gas in the room

For this step, we'll be using the Ideal Gas Law: PV = nRT. P = pressure = 1.00 atm V = volume of the gas = 61.3084 m³ (from Step 2) T = temperature = (24 + 273.15) K = 297.15 K R = gas constant = 0.0821 L atm/mol K First, we need to convert the volume from cubic meters to liters: 61.3084 m³ × (1000 L/m³) = 61308.4 L. Now, we can rearrange the Ideal Gas Law to solve for 'n': n = PV / RT n = (1.00 atm × 61308.4 L) / (0.0821 L atm/mol K × 297.15 K) n = 2492.35 moles (total moles of gas in the room)
04

Determine the number of moles of Ni(CO)₄

The allowable concentration of Ni(CO)₄ is 1 ppb, which means 1 mole of Ni(CO)₄ for every \(10^9\) moles of gas. To find the number of moles of Ni(CO)₄ in the room, use the following proportion: 1 mole Ni(CO)₄ / \(10^9\) moles of gas = x moles Ni(CO)₄ / 2492.35 moles of gas Solving for 'x': x = (1 mole Ni(CO)₄ × 2492.35 moles of gas) / \(10^9\) moles of gas x = 2.49235 × \(10^{-6}\) moles Ni(CO)₄
05

Convert moles of Ni(CO)₄ to mass

To find the mass of Ni(CO)₄, use the molar mass of Ni(CO)₄: Molar mass of Ni(CO)₄ = (Ni = 58.69 g/mol) + (4 × (C = 12.01 g/mol + O = 16.00 g/mol)) Molar mass of Ni(CO)₄ = 58.69 + 4 × (12.01 + 16.00) = 170.33 g/mol Now multiply the number of moles of Ni(CO)₄ by its molar mass to find the mass: Mass = 2.49235 × \(10^{-6}\) moles × 170.33 g/mol = 0.42446 g The allowable mass of Ni(CO)₄ in the laboratory room under the given conditions is approximately 0.42446 grams.

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Most popular questions from this chapter

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