Question

In order to determine the rate of photosynthesis (the conversion by plants of carbon dioxide and water into glucose and oxygen), the oxygen gas emitted by an aquatic plant is collected over water at a temperature of 293 K and a total pressure of 754.8 mmHg. Over a specific time-period, a total of 1.31 L of gas is collected. The partial pressure of water at 293 K is 17.55 mmHg. What mass of oxygen gas (in grams) forms

Answer 1

Answer:

1.6896 grams

Explanation:

Given that:

Total pressure = The pressure of  $O_2$ + partial pressure of $H_2O$

∴

The pressure of  $O_2$  = Total pressure - partial pressure of $H_2O$

= (754.8 - 17.55) mmHg

= 737.25 mmHg

At standard conditions:

1 mmHg = 0.00131579 atm

∴

737.25 mmHg will be: (737.25*0.00131579) atm = 0.970065928 atm

P ≅ 0.97 atm

The temperature  (T) = 293 K

Volume (V) = 1.31 L

Using ideal gas equation

PV = nRT

0.97 × 1.31 = n × 0.0821 × 293

$n = \dfrac{0.97 \times 1.31}{0.0821 \times 293}$

n = 0.0528

mass of oxygen  $O_2$ = no. of moles * molar mass

= 0.0528 * 32

= 1.6896 grams