Question

2619. A gaseous mixture of 2 moles of A, 3 moles of B,5 molesof C and 10 moles of D is contained in a vessel. Assumingthat gases are ideal and the partial pressure of C is 1.5 atm,the total pressure is:(a) 3 atm(b) 6 atm(c) 9 atm(d) 15 atm give answer step wise​

Answer 1

Explanation:

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Answer 2

The correct answer is option b.

Explanation:

Total pressure of the gases = P

Moles of gas A = 2 mol

Moles of gas B = 3 moles

Moles of gas C = 5 mol

Moles of gas D = 10 mol

Partial pressure of the gas C = $p_c=1.5 atm[/tex</p><p>Mole fraction of gas C = [tex]\chi_c$

$\chi_c=\frac{\text{Moles of gas C}}{\text{um of all moles of gases}}$

$\chi_c=\frac{5 mol}{2mol+3mol+5mol+10mol}=0.25$

According to Raoult's law;

$p_i=P\times \chi_i$

$p_c=P\times \chi_c$

$P=\frac{p_c}{\chi_c}=\frac{1.5 atm}{0.25}=6 atm$

The correct answer is option b.

Learn more about : Raoult's law

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