NEET प्रश्न (MCQ) हिंदी में — उत्तर और विस्तृत हल सहित

$$\begin{gathered} \left(\mathrm{D}\right). {\mathrm{r}}_{{\mathbf{H}}_{\mathbf{2}}}=\frac{{\mathrm{V}}_{{\mathbf{H}}_{\mathbf{2}}}}{\mathrm{t}} ; {\mathrm{r}}_{{\mathbf{O}}_{\mathbf{2}}}=\frac{{\mathrm{V}}_{{\mathbf{O}}_{\mathbf{2}}}}{\mathrm{t}} \\ \therefore \frac{ {\mathrm{r}}_{{\mathbf{H}}_{\mathbf{2}}}}{{\mathrm{r}}_{{\mathbf{O}}_{\mathbf{2}}}}=\frac{{\mathrm{V}}_{{\mathbf{H}}_{\mathbf{2}}}}{{\mathrm{V}}_{{\mathbf{O}}_{\mathbf{2}}}}=\frac{1}{{\mathrm{n}}_{{\mathbf{O}}_{\mathbf{2}}}} \\ \mathrm{या} \frac{{\mathrm{r}}_{{\mathbf{O}}_{\mathbf{2}}}}{{\mathrm{r}}_{{\mathbf{H}}_{\mathbf{2}}}}={\mathrm{n}}_{{\mathbf{O}}_{\mathbf{2}}}=\frac{1}{\sqrt{16}}=\frac{1}{4} \\ \therefore {\mathrm{m}}_{{\mathbf{O}}_{\mathbf{2}}}=\frac{1}{4}\times 32=8 \mathrm{ग्राम}. \end{gathered}$$

$$\begin{gathered} \left(\mathrm{A}\right). \mathrm{The} \mathrm{mean} \mathrm{free} \mathrm{path} \mathrm{of} \mathrm{gas} \mathrm{molecules}, \\ \mathcal{l}=\frac{1}{\sqrt{2}{\mathrm{\pi n\sigma }}^2}; \mathcal{l}\propto \frac{1}{{\mathrm{n\sigma }}^2}; \mathcal{l}\propto \frac{1}{{\mathrm{P\sigma }}^2} \\ 6990 \mathrm{cm} \propto \frac{1}{{10}^{-6}\times 4^2}; \mathrm{x} \mathrm{cm} \propto \frac{1}{{10}^{-3}\times 2^2} \\ \frac{6690}{\mathrm{x}}=\frac{{10}^6}{16}\times {10}^{-3}\times 4=250 \\ \therefore \mathrm{x}=\frac{6990}{250}=27.96 \mathrm{cm}\approx 28 \mathrm{cm}. \end{gathered}$$

(C).  अणुओं की समान संख्या, इसका अर्थ है गैस के मोलों की समान संख्या उपस्थित है।

$$\begin{gathered} \therefore {\mathrm{n}}_{{\mathbf{O}}_{\mathbf{2}}}={\mathrm{n}}_{{\mathbf{N}}_{\mathbf{2}}}=\mathrm{x} \mathrm{say} \\ \therefore {\mathrm{P}}_1\mathrm{V}=\left({\mathrm{n}}_{{\mathbf{O}}_{\mathbf{2}}}+{\mathrm{n}}_{{\mathbf{N}}_{\mathbf{2}}}\right)\mathrm{RT}=2\times \mathrm{RT} \\ {\mathrm{P}}_2\mathrm{V}={\mathrm{n}}_{{\mathbf{O}}_{\mathbf{2}}\mathbf{ }}\mathrm{RT}=\mathrm{x} \mathrm{RT} \\ \therefore \frac{{\mathrm{P}}_1}{{\mathrm{P}}_2}=2 \mathrm{or} {\mathrm{P}}_1=2{\mathrm{P}}_2 \\ \therefore \mathrm{when} \mathrm{nitrogen} \mathrm{is} \mathrm{removed}, \mathrm{final} \mathrm{pressure} \mathrm{will} \mathrm{be} \mathrm{P}/2. \end{gathered}$$

$\left(\mathrm{D}\right). \frac{{\mathrm{n}}_{\mathrm{f}}}{{\mathrm{n}}_{\mathrm{i}}}=\frac{{\mathrm{P}}_{\mathrm{f}}}{{\mathrm{P}}_{\mathrm{i}}} ; {\mathrm{P}}_{\mathrm{f}}={\mathrm{P}}_{\mathrm{i}}\frac{{\mathrm{n}}_{\mathrm{f}}}{{\mathrm{n}}_{\mathrm{i}}} \mathrm{या} \left(2\right) \left(\frac{3}{2}\right)=3 \mathrm{atm}$

$$\begin{gathered} \left(\mathrm{C}\right). 32 \mathrm{gm} {\mathrm{O}}_2=1 \mathrm{mole} {\mathrm{O}}_2 \\ 3 \mathrm{gm} \mathrm{hydrogen}=\frac{3}{2}=1.5 \mathrm{mole} {\mathrm{H}}_2 \\ \mathrm{Hence} \mathrm{total} \mathrm{moles} \mathrm{of} \mathrm{gas} \mathrm{present}=2.5 \\ \therefore \mathrm{volume} \mathrm{of} \mathrm{total} 2.5 \mathrm{moles} \mathrm{of} \mathrm{gas} \mathrm{mix} \mathrm{at} \mathrm{STP} \\ =2.5\times 22.4 = 56 \mathrm{lit}. \end{gathered}$$

(C). गैस का ‘a’ मान जितना अधिक होगा, गैस के अणुओं के बीच अंतर-आण्विक आकर्षण उतना ही अधिक होगा,

       इसलिए द्रवीकरण उतना ही आसान होगा।

$$\begin{gathered} \left(\mathrm{B}\right). \mathrm{K}.\mathrm{E}.=\frac{3}{2}\mathrm{RT} \\ \therefore {\left(\mathrm{K}.\mathrm{E}.\right)}_1=\frac{3}{2}\times \mathrm{R}\times 400; {\left(\mathrm{K}.\mathrm{E}.\right)}_2=\frac{3}{2}\times \mathrm{R}\times 800 \\ \therefore \frac{{\left(\mathrm{K}.\mathrm{E}.\right)}_2}{{\left(\mathrm{K}.\mathrm{E}.\right)}_1}=2 \mathrm{या} {\left(\mathrm{K}.\mathrm{E}.\right)}_2=2{\left(\mathrm{K}.\mathrm{E}.\right)}_1 \end{gathered}$$

$$\begin{gathered} \left(\mathrm{A}\right). \mathrm{K}.\mathrm{E}. \mathrm{for} {\mathrm{UF}}_6=\frac{1}{2}\times {\mathrm{M}}_{{\mathbf{UF}}_{\mathbf{6}}}\times \frac{3\mathrm{RT}}{{\mathrm{M}}_{{\mathbf{UF}}_{\mathbf{6}}}}=\frac{3}{2}\mathrm{RT} \\ \mathrm{K}.\mathrm{E}. \mathrm{for} {\mathrm{H}}_2=\frac{1}{2}\times {\mathrm{M}}_{{\mathbf{H}}_{\mathbf{2}}}\times \frac{3\mathrm{RT}}{{\mathrm{M}}_{{\mathbf{H}}_{\mathbf{2}}}}=\frac{3}{2}\mathrm{RT} \\ \therefore \mathrm{K}.\mathrm{E}. \mathrm{of} {\mathrm{UF}}_6 \mathrm{to} \mathrm{that} {\mathrm{H}}_2 \mathrm{is} 1 \mathbf{:}\mathbf{ }1. \end{gathered}$$

$$\begin{gathered} \left(\mathrm{B}\right). {\mathrm{r}}_{{\mathbf{H}}_{\mathbf{2}}}=\frac{\mathrm{x}}{5} {\mathrm{r}}_{\mathbf{x}}=\frac{\mathrm{x}}{\mathrm{t}} \\ \therefore \frac{\mathrm{x}}{5}\times \frac{\mathrm{t}}{\mathrm{x}}=\sqrt{\frac{{\mathrm{M}}_{\mathbf{x}}}{{\mathrm{M}}_{{\mathbf{H}}_{\mathbf{2}}}}} \\ \mathrm{or} \mathrm{t}=\sqrt{\frac{{\mathrm{M}}_{\mathbf{x}}}{{\mathrm{M}}_{{\mathbf{H}}_{\mathbf{2}}}}}\times 5 \mathrm{or} \mathrm{t}=\sqrt{\frac{{\mathrm{M}}_{\mathbf{x}}}{2}}\times 5 \\ \mathrm{For} \mathrm{He} : {\mathrm{M}}_{\mathrm{x}}=4, \mathrm{t}=5\sqrt{2} \\ \mathrm{For} \mathrm{CO} : {\mathrm{M}}_{\mathrm{x}}=28, \mathrm{t}=5\sqrt{14} \\ \mathrm{For} {\mathrm{O}}_2 : {\mathrm{M}}_{\mathrm{x}}=32, \mathrm{t}=20 \\ \mathrm{For} {\mathrm{CO}}_2 : {\mathrm{M}}_{\mathrm{x}}=44, \mathrm{t}=5\sqrt{22} \end{gathered}$$