Question

Oxygen gas generated by the decomposition of Potassium chlorate is collected over Water the volume of oxygen collected at 24 degree Celsius and atmospheric pressure of 760 mm of HG is 128 ml calculate the mass of Oxygen gas obtained the pressure of the water vapour at 25 degree Celsius is 22.4 mm​

Answer 1

Answer:

The mass of oxygen gas obtained is 0.1561 grams.

Explanation:

Atmospheric pressure = 760 mmHg

Vapor pressure of the water = 22.4 mmHg

Pressure of oxygen gas on the water = p

Atmospheric pressure = Vapor pressure of the water +$p_{O_2}$

760 mmHg = 22.4 mmHg + $p_{O_2}$

$p_{O_2}=760 mmHg - 22.4 mmHg=737.6 mmHg$

Volume of oxygen gas =V = 128 mL = 0.128 L

Pressure of oxygen gas = P =$p_1=737.6 mmHg=0.97 atm$

Moles or oxygen ags = n

Temperature of the gas = T = 24°C = 297.15 K

Using an ideal gas equation:

$n=\frac{PV}{RT}$

$n=\frac{0.93 atm\times 0.128 L}{0.0821 atm L.mol K\times 297.15 K}$

n = 0.004879 mol

Mass of 0.004849 moles of oxygen gas :

0.004879 mol × 32 g/mol = 0.1561 g

The mass of oxygen gas obtained is 0.1561 grams.