Question

a solution of urea is prepared by dissolving of 180g of it one kg of water the mole fraction of urea in the solution​

Answer 1

$\textsf{\Large {\underline {Solution}}}$ :

Mass of urea, $\mathsf{NH_2CONH_2}$ = 180 g

Molar Mass of urea, $\mathsf{NH_2CONH_2}$ = 14 + 2 +12 + 16 + 14 + 2 = 60 g

Moles of urea, $\mathsf{{n} _{\tiny{NH_2CONH_2}}}$ = $\mathsf{ \dfrac{Given \:mass\:of\:urea} {Molar\:mass\:of\:urea}}$ = $\mathsf{\dfrac{180}{60}}$ = 3 mol

Given mass of water = 1kg = 1000 g

Molar Mass of water, $\mathsf{H_{2}O}$ = 2 + 16 = 18g

Moles of water, $\mathsf{{n}_{\tiny{H_{2}O}}}$ = $\mathsf{ \dfrac{Given \:mass\:of\:water} {Molar\:mass\:of\:water}}$

Moles of water = $\mathsf{\dfrac{1000}{18}}$ = 55.55 mol

Mole fraction of Urea in the solution, $\mathsf{{\chi}_{\tiny{NH_2CONH_2}}}$ = $\mathsf{\dfrac{{n} _{\tiny{NH_2CONH_2}}} {n _{\tiny{NH_2CONH_2}}\:+\:{n_{\tiny{H_{2}O}}}}}$

Mole fraction of urea, $\mathsf{{\chi}_{\tiny{NH_2CONH_2}}}$= $\mathsf{\dfrac{3}{3 \:+\:55.55}}$

Mole fraction of urea, $\mathsf{{\chi}_{\tiny{NH_2CONH_2}}}$ = $\mathsf{\dfrac{3}{58.55}}$

➡️ Mole fraction of urea, $\mathsf{{\chi}_{\tiny{NH_2CONH_2}}}$ = $\mathsf{0.0512}$

$\textsf{\Large {\underline {About Mole Fraction}}}$ :

Mole fraction of any component in solution is the ratio of number of moles of that component to the total number of moles of all the components present in the solution.

$\textsf{\large{\underline {Formula}}}$ :

$\boxed{\mathsf{\chi} _{\tiny{A}} = {\dfrac{n_{\tiny{A}}} {n_{\tiny{A}} \:+\:n_{\tiny{B}}}}}$

$\boxed{\mathsf{\chi} _{\tiny{B}} = {\dfrac{n_{\tiny{B}}} {n_{\tiny{A}} \:+\:n_{\tiny{B}}}}}$

⚫ The sum of all components of a mole fraction is always equal to 1.

Example : $\mathsf{{\chi} _{\tiny{A}} + {\chi} _{\tiny{B}} = 1}$