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Question
use the following information to answer the next question.
a scientist isolates 2.366 g of gas. the sample occupies a volume of 800 ml at 78.0°c and 103 kpa. the scientist uses this data to calculate the molar mass and thus identify the gas.
- the gas the scientist isolated was
xe
f₂
br₂
kr
Step1: Convert units
- Convert temperature from Celsius to Kelvin: \(T=(78.0 + 273.15)\text{K}=351.15\text{K}\)
- Convert volume from mL to L: \(V = 800\text{mL}=0.8\text{L}\)
- Use the ideal gas law \(PV=nRT\), where \(P = 103\text{kPa}\), \(R=8.314\text{J/(mol}\cdot\text{K)}\) (since \(1\text{kPa}\cdot\text{L}=1\text{J}\))
- Rearrange for \(n\): \(n=\frac{PV}{RT}\)
$$n=\frac{103\times0.8}{8.314\times351.15}$$
$$n=\frac{82.4}{2920.0}$$
$$n\approx0.0282\text{mol}$$
Step2: Calculate molar mass
- Molar mass \(M=\frac{m}{n}\), where \(m = 2.366\text{g}\)
$$M=\frac{2.366}{0.0282}$$
$$M\approx84\text{g/mol}$$
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Kr (since the molar mass of Kr is approximately \(83.8\text{g/mol}\))