Answered

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C. It is desired to obtain a volume of 1.00 L of nitrogen
at 112.0 °C and 660. mmHg. How many grams of
nitrogen would be required?


Sagot :

SOLUTION:

Step 1: List the given values.

[tex]\begin{aligned} & P = \text{660. mmHg = 0.86842 atm} \\ & V = \text{1.00 L} \\ & T = 112.0^{\circ}\text{C} = \text{385.15 K} \end{aligned}[/tex]

Step 2: Calculate the number of moles of gas using ideal gas equation.

[tex]\begin{aligned} n & = \frac{PV}{\text{RT}} \\ & = \frac{(\text{0.86842 atm})(\text{1.00 L})}{\left(0.082057 \: \dfrac{\text{L}\cdot\text{atm}}{\text{mol}\cdot\text{K}}\right)(\text{385.15 K})} \\ & = \text{0.0275 mol} \end{aligned}[/tex]

Step 3: Calculate the mass of the gas.

The molar mass of N₂ is 28.0 g/mol.

[tex]\begin{aligned} \text{mass} & = n \times \text{MM} \\ & = \text{0.0275 mol} \times \text{28.0 g/mol} \\ & = \boxed{\text{0.770 g}} \end{aligned}[/tex]

Hence, 0.770 g of nitrogen would be required.

[tex]\\[/tex]

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