Question

7. The mole fraction of nitrogen in a mixture
containing 70 gm nitrogen, 120 gm of oxygen
and 44 gm of CO2 is
(A) 0.36 (B) 0.34 (C) 0.29 (D) 5.0​

Answer 1

$\Large\bf{\color{blue}{\underline{\underline{Given:-}}}}$

$\bf Weight \:of\: nitrogen = 70g$

$\bf Weight\: of \:oxygen = 120 g$

$\bf Weight\: of\: carbon\: dioxide = 44 g$

$\Large\bf{\color{blue}{\underline{\underline{To\:Find:-}}}}$

$\bf Molecular\: fraction\:of\:nitrogen$

$\Large\bf{\color{blue}{\underline{\underline{Solution:-}}}}$

$\bf{\underline {\boxed{\pink{Moles \: = \dfrac{Weight}{Molecular \:weight}}}}}$

$\bf Moles\: of \:N₂(n₁)= \dfrac{70}{28} = 2.5$

$\bf moles \:of\: O₂(n₂) = \dfrac{120}{32} = 3.75$

$\bf moles\: of \: CO₂(n₃) = \dfrac{44}{44} = 1$

$\bf \therefore \:mole\: fraction \:of\: nitrogen(N₂)\\ \bf :\implies \dfrac{n₁}{ n₁+n₂+n₃ }$

$\bf:\implies \dfrac{2.5}{ 2.5+3.75+1}$

$\bf:\implies \dfrac{2.5}{7.25} =0.34$

$\bf {\purple {\boxed{Option (B)=0.34}}}$