Question

the total pressure of 6.4 grams of oxygen and 5.6 grams of nitrogen present in a 2 lit vessel is 1200 mm What is the partial pressure of nitrogen in mm​

Answer 1

moles of nitrogen = 5.6/14 = 0.4
moles of oxygen = 6.4/16 = 0.4
mole fraction of nitrogen = 0.4/0.8 = 0.5

so, partial pressure of nitrogen is

0.5x1200 = 600

Answer 2

The partial pressure of nitrogen in mm​ is 600

Explanation]

Using the equation given by Raoult's law, we get:

$p_A=\chi_A\times P_T$

$p_{N_2}$ = partial pressure of $N_2$ = ?

$\chi_{N_2}$ = mole fraction of $N_2$ =

$P_{T}$ = total pressure of solution = 1200 mm

Mole fraction of nitrogen is written as:

$\chi_{N_2}=\frac{\text{Moles of }N_2}{\text{Total moles}}$

$\text{Number of moles of nitrogen}=\frac{\text{Given mass}}{\text{Molar mass}}=\frac{5.6g}{28g/mol}=0.2mol$

$\text{Number of moles of oxygen}=\frac{\text{Given mass}}{\text{Molar mass}}=\frac{6.4g}{32/mol}=0.2mol$

$\chi_{N_2}=\frac{0.2}{0.2+0.2}=0.5$

$p_{N_2}=0.5\times 1200=600mm$

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