Answer: The partial pressure in atm of ammonium telluride is 1.48 atm
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
[tex]p_A=x\times p_{total}[/tex]
where, x = mole fraction
[tex]p_A[/tex] = partial pressure of A
[tex]p_{total}[/tex] = total pressure
[tex]x_{\text {ammonium telluride}}=\frac{\text {moles of ammonium telluride}}{\text {total moles}}[/tex],
[tex]x_{\text {ammonium telluride}}=\frac{\frac{2.62g}{164g/mol}}{\frac{2.62g}{164g/mol}+\frac{3.83g}{333g/mol}+\frac{1.82g}{673g/mol}}=0.53[/tex]
[tex]p_{\text {ammonium telluride}}=0.53\times 2.795atm=1.481atm[/tex]
Thus the partial pressure in atm of ammonium telluride is 1.48 atm
