Question

A tank consists of a gaseous mixture of 3.83 g of triselenium diiodide, 2.62 g of ammonium telluride, and 1.82 g of iridium (i) sulfate. What is the partial pressure in atm of ammonium telluride if the total pressure in the tank is 2.795 atm?

Answer 1

Answer: The partial pressure in atm of ammonium telluride is 1.48 atm

Step-by-step explanation:

According to Raoult's law, the partail pressure of a component at a given temperature is equal to the mole fraction of that component multiplied by the total pressure

`p_A=x* p_(total)`

where, x = mole fraction

`p_A` = partial pressure of A

`p_(total)` = total pressure

`x_{\text {ammonium telluride}}=\frac{\text {moles of ammonium telluride}}{\text {total moles}}`,

`x_{\text {ammonium telluride}}=((2.62g)/(164g/mol))/((2.62g)/(164g/mol)+(3.83g)/(333g/mol)+(1.82g)/(673g/mol))=0.53`

`p_{\text {ammonium telluride}}=0.53* 2.795atm=1.481atm`

Thus the partial pressure in atm of ammonium telluride is 1.48 atm