ଏକକ 2 ସମାଧାନ (ଅନ୍ତର୍ଗତ ପ୍ରଶ୍ନ-2)

ଉତ୍ତର

ଏହା ଦିଆଯାଇଛି ଯେ STP ରେ ଜଳରେ ${H_2} {S}$ ର ଦ୍ରାବ୍ୟତା $0.195 {~m}$ ଅଟେ ଅର୍ଥାତ୍, ${H_2} {S}$ ର 0.195 ମୋଲ $1000 {~g}$ ଜଳରେ ଦ୍ରବୀଭୂତ ହୁଏ।

ଜଳର ମୋଲ ସଂଖ୍ୟା $=\dfrac{1000 {~g}}{18 {~g} {~mol}^{-1}}$ $=55.56 {~mol}$

$$ \text {Mole fraction of } H_2 S =\dfrac{\text { Moles of } {H_2} {S}}{\text { Moles of } {H_2} {S}+\text { Moles of water }} $$

$$\quad\quad\quad\quad\quad =\dfrac{0.195}{0.195+55.56} =0.0035 $$

STP ରେ, ଚାପ $(p)=0.987$ ବାର

ହେନ୍ରୀଙ୍କ ନିୟମ ଅନୁଯାୟୀ : $ \quad p={K_{H}} \chi$

$ \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad \Rightarrow {K_{H}}=\dfrac{p}{\chi}$

$ \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad \quad \quad \quad =\dfrac{0.987}{0.0035} \hspace{0.5mm}bar $

$ \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad \quad \quad \quad = 282 \hspace{0.5mm}bar $

ଉତ୍ତର

ଏହା ଦିଆଯାଇଛି ଯେ :

${K_{H}}=1.67 \times 10^{8} {~Pa}$

$ p_{{CO_2}}=2.5 {~atm}=2.5 \times 1.01325 \times 10^{5} {~Pa} $

$\quad\quad\quad\quad\quad\quad\quad=2.533125 \times 10^{5} {~Pa}$

ହେନ୍ରୀଙ୍କ ନିୟମ ଅନୁଯାୟୀ:

$$ \begin{aligned} p_{{CO_2}} & ={K_{H}} \chi \\ \Rightarrow \chi & =\dfrac{p_{{CO_2}}}{{~K_{H}}} \\ & \chi =\dfrac{2.533125 \times 10^{5}}{1.67 \times 10^{8}} \end{aligned} $$

$$\Rightarrow \chi=0.00152$$

ଆମେ ଲେଖିପାରିବା,

$$ \chi =\dfrac{n_{{CO_2}}}{n_{{CO_2}}+n_{{H_2} {O}}} \approx \dfrac{n_{{CO_2}}}{n_{{H_2} {O}}} $$

[କାରଣ, $n_{{CO_2}}$ ହେଉଛି ${n_{{H_2} {O}}}$ ତୁଳନାରେ ନଗଣ୍ୟ]

$500 {~mL}$ ସୋଡା ଜଳରେ, ଜଳର ଆୟତନ $=500 {~mL}$ [ସୋଡାର ପରିମାଣକୁ ଅଣଦେଖା କରି]

ଆମେ ଲେଖିପାରିବା:

$500 {~mL}$ ଜଳ $=500 {~g}$ ଜଳ

$\quad\quad\quad\quad\quad\quad\quad=\dfrac{500}{18} {~mol}$ ଜଳ

$\quad\quad\quad\quad\quad\quad\quad=27.78 {~mol}$ ଜଳ

ବର୍ତ୍ତମାନ, $\quad\dfrac{n_{{CO_2}}}{n_{{H_2} {O}}}=\chi$

$\quad\quad \quad\dfrac{n_{{CO_2}}}{27.78}=0.00152$

$n_{{CO_2}}=0.042 {~mol}$

ତେଣୁ, $500 {~mL}$ ସୋଡା ଜଳରେ ${CO_2}$ ର ପରିମାଣ $=(0.042 \times 44) {g}$

$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad=1.848 {~g}$