Question

A 12400. mL container holds a sample of argon gas at 35.00C and

Answer 1

Answer:

0.574moles

Explanation:

Using the general gas equation;

PV = nRT

Where;

P = pressure (atm)

V = volume (Litres)

n = number of moles (mol)

R = gas law constant (0.0821 Latm/molK)

T = temperature (Kelvin)

According to the information provided in the question;

- Volume (V) = 12400mL = 12400/1000 = 12.4L

- Pressure (P) = 890mmHg = 890/760 = 1.17atm

- Temperature (T) = 35°C = 35 + 273 = 308K

Hence, using PV = nRT

n = PV/RT

n = 1.17 × 12.4 ÷ 0.0821 × 308

n = 14.508 ÷ 25.287

n = 0.574moles

Therefore, the number of moles of argon gas in the cylinder is 0.574moles