Question

calculate to three significant digits the density of dinitrogen monoxide gas at exactly 30°c and exactly 1 atm. you can assume dinitrogen monoxide gas behaves as an ideal gas under these conditions.

Explanation:

Step1: Convert temperature to Kelvin

$$T = 30 + 273.15 = 303.15\ K$$

Step2: Find molar mass of \(N_2O\)

Molar mass of \(N = 14.01\ g/mol\), molar mass of \(O=16.00\ g/mol\).
$$M=(2\times14.01 + 16.00)\ g/mol=44.02\ g/mol$$

Step3: Use ideal gas law \(PV = nRT\) and \(n=\frac{m}{M}\), \(d=\frac{m}{V}\)

From \(PV=\frac{m}{M}RT\), we can derive \(d=\frac{m}{V}=\frac{PM}{RT}\)
Given \(P = 1\ atm\), \(R=0.0821\ L\cdot atm/(mol\cdot K)\), \(M = 44.02\ g/mol\), \(T = 303.15\ K\)
$$d=\frac{1\ atm\times44.02\ g/mol}{0.0821\ L\cdot atm/(mol\cdot K)\times303.15\ K}$$

Step4: Calculate the density

$$d=\frac{44.02}{0.0821\times303.15}\ g/L$$
$$0.0821\times303.15 = 24.888615$$
$$d=\frac{44.02}{24.888615}\ g/L\approx1.77\ g/L$$

Answer:

\(1.77\)