Answer to Question #156731 in General Chemistry for tej

Question

Halothane (CHBrClF) is used as an anesthetic. How many grams of halothane would need to be administered to completely fill the lungs of an average adult human, given an average lung volume of 5.95 L and a body temperature of 37°C? Assume a pressure of 1.00 atm.

Answer 1

V = 5.95L = 5.95 × 10^-3 m³

T = 37°C = 273 + 37 = 310K

P = 1.00 atm = 101325 Nm^-2

Using the general gas law;

PV = nRT

101325 × 5.95 × 10^-3 = n × 8.314 × 310

n = $\dfrac{101325 × 5.95 × 10^{-3 }}{8.314 × 310}$ = 0.234 moles

1 mole of halothane = 147.37g

0.234 mole of halothane = ? = 0.234 × 147.37 = 34.48g

Therefore, 34.48 grams of halothane would need to be administered.