Answer:
7.13 L of ethanol, CH₃CH₂OH
Explanation:
We'll begin by calculating the number of mole in 2.24 g of ethanol. This can be obtained as follow:
Mass of CH₃CH₂OH = 2.24 g
Molar mass of CH₃CH₂OH = 12 + (3×1) + 12 + (2×1) + 16 + 1
= 12 + 3 + 12 + 2 + 16 + 1
= 46 g/mol
Mole of CH₃CH₂OH =?
Mole = mass / molar mass
Mole of CH₃CH₂OH = 2.24 / 46
Mole of CH₃CH₂OH = 0.049 mole
Next, we shall convert 40 °C to Kelvin temperature. This can be obtained as follow:
T(K) = T(°C) + 273
T(°C) = 40 °C
T(K) = 40 °C + 273
T(K) = 313 K
Finally, we shall determine the volume of ethanol, CH₃CH₂OH in present in the container. This can be obtained as follow:
Pressure (P) = 17.88 KPa
Temperature (T) = 313 K
Number of mole (n) = 0.049 mole
Gas constant (R) = 8.314 L.KPa/Kmol
Volume (V) =?
PV = nRT
17.88 × V = 0.049 × 8.314 × 313
Divide both side by 17.88
V = (0.049 × 8.314 × 313) / 17.88
V = 7.13 L
Thus, 7.13 L of ethanol, CH₃CH₂OH will be present in the container.
