We are asked** how much liquid butane is present** in a **300. mL sealed flask contains 0.620g of butane at -21.0°C**

**We go through the following steps to solve the problem: **

**Step 1. **Calculate the vapor pressure at -21.0 °C

**Step 2.** Calculate the mass in the gas phase

**Step 3.** Calculate the mass in the liquid phase

**Step 1. **Calculate the vapor pressure at -21.0 °C

For this part, we can use the ** Clausius-Clapeyron Equation**:

$\overline{){\mathbf{ln}}\frac{{\mathbf{P}}_{\mathbf{2}}}{{\mathbf{P}}_{\mathbf{1}}}{\mathbf{}}{\mathbf{=}}{\mathbf{}}{\mathbf{-}}\frac{\mathbf{\u2206}{\mathbf{H}}_{\mathbf{vap}}}{\mathbf{R}}{\mathbf{[}}\frac{\mathbf{1}}{{\mathbf{T}}_{\mathbf{2}}}{\mathbf{-}}\frac{\mathbf{1}}{{\mathbf{T}}_{\mathbf{1}}}{\mathbf{]}}}$

where:

**P _{1}** = vapor pressure at T

**P _{2}** = vapor pressure at T

**ΔH _{vap}** = heat of vaporization (in J/mol)

**R** = gas constant (8.314 J/mol•K)

**T _{1} and T_{2}** = temperature (in K).

Butane (C_{4}H_{10}) has a heat of vaporization of 22.44 kJ/mol and a normal boiling point of -0.40°C. A 300. mL sealed flask contains 0.620g of butane at -21.0°C. How much liquid butane is present?

A. 0.375g

B. 0.475g

C. 0.245g

D. 0.620g

E. no liquid left

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