NEET Practice Questions (MCQs) with Answers & Solutions

$$\begin{gathered} \left(\mathrm{B}\right). \mathrm{For} \mathrm{ideal} \mathrm{gases}. \\ \mathrm{PV}=\mathrm{nRT}=\frac{\mathrm{m}}{\mathrm{M}} \mathrm{RT}:\mathrm{P}=\frac{\mathrm{RT}}{\mathrm{M}}\left(\frac{\mathrm{m}}{\mathrm{V}}\right)=\frac{\mathrm{RT}}{\mathrm{M}} \mathrm{d} \\ \left(\mathrm{or}\right) \mathrm{M}=\mathrm{RT}\left(\frac{\mathrm{d}}{\mathrm{P}}\right). \\ \mathrm{Given} : \mathrm{d}=2.00 \mathrm{P}+0.020 {\mathrm{P}}^2\left(\mathrm{for} \mathrm{a} \mathrm{real} \mathrm{gas}\right). \\ \frac{\mathrm{d}}{\mathrm{P}}=2.00+0.040 \mathrm{P} : \underset{\mathrm{P}\to 0}{\mathrm{Lt}}\frac{\mathrm{d}}{\mathrm{P}}=2.00, \\ \mathrm{which} \mathrm{is} \frac{\mathrm{d}}{\mathrm{P}} \mathrm{for} \mathrm{an} \mathrm{ideal} \mathrm{gas}. \\ \mathrm{Thus} \mathrm{M}=\mathrm{RT} \mathrm{x} 2=25 \mathrm{x} 2=50 \mathrm{g} {\mathrm{Mol}}^{-1}. \end{gathered}$$

$$\begin{gathered} \left(\mathrm{C}\right). \mathrm{Mass} \mathrm{of} \mathrm{water} \mathrm{filled} \mathrm{in} \mathrm{the} \mathrm{glass} \mathrm{vessel}, \\ {\mathrm{m}}_1=\left(148.0-50.0\right)\mathrm{g} =98.0 \mathrm{g} \\ \mathrm{Volume} \mathrm{of} \mathrm{glass} \mathrm{vessel}, \\ \mathrm{V}=\frac{{\mathrm{m}}_1}{\mathrm{\rho }}=\frac{98.0 \mathrm{g}}{0.98 \mathrm{g} {\mathrm{mol}}^{-1}}= 100 \mathrm{mL} = 0.1 {\mathrm{dm}}^3 \\ \mathrm{Mass} \mathrm{of} \mathrm{gas} \mathrm{filled} \mathrm{in} \mathrm{the} \mathrm{vessel}, \mathrm{m} = \left(50.5 – 50.0\right)\mathrm{g} = 0.5 \mathrm{g} \\ \mathrm{If} \mathrm{M} \mathrm{is} \mathrm{the} \mathrm{molar} \mathrm{mass} \mathrm{of} \mathrm{the} \mathrm{gas}, \mathrm{we} \mathrm{will} \mathrm{have} \\ \mathrm{pV}=\mathrm{nRT}=\frac{\mathrm{m}}{\mathrm{M}}\mathrm{RT} \\ \mathrm{or} \mathrm{M}=\frac{\mathrm{mRT}}{\mathrm{pV}} \\ =\frac{\left(0.5\mathrm{g}\right)\left(8.314 \mathrm{J} {\mathrm{K}}^{-1} {\mathrm{mol}}^{-1}\right)\left(300 \mathrm{K}\right)}{\left(101.325 \mathrm{kPa}\right)\left(0.1 {\mathrm{dm}}^3\right)}=123 \mathrm{g} {\mathrm{mol}}^{-1} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{A}\right). \mathrm{We} \mathrm{have} {\mathrm{u}}_{\mathrm{av}}=\sqrt{\frac{8\mathrm{RT}}{\mathrm{\pi M}}} \\ \mathrm{Hence}, \sqrt{\frac{8{\mathrm{RT}}_1}{\mathrm{\pi M}}}=9.0\times {10}^4 \mathrm{cm} {\mathrm{s}}^{-1} \\ \mathrm{or} {\mathrm{T}}_1={\left(9.0\times {10}^2 \mathrm{m} {\mathrm{s}}^{-1}\right)}^2\left(\frac{\mathrm{\pi M}}{8\mathrm{R}}\right)={\left(9.0\times {10}^2 \mathrm{m} {\mathrm{s}}^{-1}\right)}^2 \\ \left[\frac{\left(3.14\right)\left(44\times {10}^{-3} \mathrm{kg} {\mathrm{mol}}^{-1}\right)}{\left(8\right)\left(8.314 \mathrm{J} {\mathrm{K}}^{-1} {\mathrm{mol}}^{-1}\right)}\right]=1682.5 \mathrm{K} \\ \mathrm{For} \mathrm{most} \mathrm{probable} \mathrm{speed}, \mathrm{we} \mathrm{have} \\ \sqrt{\frac{3{\mathrm{RT}}_2}{\mathrm{M}}}=9.0\times {10}^2 \mathrm{m} {\mathrm{s}}^{-1} \\ \mathrm{Hence}, {\mathrm{T}}_2={\left(9.0\times {10}^2 \mathrm{m} {\mathrm{s}}^{-1}\right)}^2\left(\frac{\mathrm{M}}{2\mathrm{R}}\right) \\ ={\left(9.0\times {10}^2 \mathrm{m} {\mathrm{s}}^{-1}\right)}^2\left[\frac{44\times {10}^{-3} \mathrm{kg} {\mathrm{mol}}^{-1}}{\left(2\right)\left(8.314 \mathrm{J} {\mathrm{K}}^{-1} {\mathrm{mol}}^{-1}\right)}\right]=2143.4 \mathrm{K} \end{gathered}$$

(B). Each flask initially contains 0.35 mole of ${\mathrm{H}}_2$

      Let ‘x’ moles of hydrogen gas be diffused from flask II to flask I

      No. of moles of ${\mathrm{H}}_2$ in flask I = (0.35 + x)

      No. of moles of ${\mathrm{H}}_2$ in flask II = (0.35 – x)

      If the new pressure is P, then

      In flask I, PV = (0.35 + x) × R × 300

      In flask II, PV = (0.35 – x) × R × 400 $\Rightarrow$x = 0.05

      If volume of each flask is ‘V’ litre $\Rightarrow$V = 17.241 L

      So, P × 17.241 = 0.30 × 0.0821 × 400$\Rightarrow$P = 0.5714 atm

(A). Let the volume of ${\mathrm{CH}}_4$ at N.T.P. = x ml

       Let the volume of ${\mathrm{C}}_2{\mathrm{H}}_4$ at N.T.P. = y ml

       Let the volume of ${\mathrm{C}}_2{\mathrm{H}}_2$ at N.T.P. = z ml

       From question, x + y + z = 10 ............ (1)

       As we know that

       Weight of ${\mathrm{CH}}_4$ + Weight of ${\mathrm{C}}_2{\mathrm{H}}_4$ + Weight of ${\mathrm{C}}_2{\mathrm{H}}_2$ = Weight of mixture

$\frac{16\mathrm{x}}{22400}+\frac{28\mathrm{x}}{22400}+\frac{262}{22400}=\frac{11.30}{11200}\times 10$ .....................(2)

$$\begin{gathered} \mathrm{Now}, \underset{\mathrm{x} \mathrm{ml}}{{\mathrm{CH}}_4}\left(\mathrm{g}\right)+\underset{2\mathrm{x} \mathrm{ml}}{2{\mathrm{O}}_2}\left(\mathrm{g}\right) \rightleftharpoons {\mathrm{CO}}_2\left(\mathrm{g}\right)+2{\mathrm{H}}_2\mathrm{O}\left(\mathcal{l}\right) \\ \underset{\mathrm{y} \mathrm{ml}}{{\mathrm{C}}_2{\mathrm{H}}_4}\left(\mathrm{g}\right)+\underset{3\mathrm{y} \mathrm{ml}}{3{\mathrm{O}}_2}\left(\mathrm{g}\right) \rightleftharpoons 2{\mathrm{CO}}_2\left(\mathrm{g}\right)+2{\mathrm{H}}_2\mathrm{O}\left(\mathcal{l}\right) \\ \underset{\mathrm{z} \mathrm{ml}}{{\mathrm{C}}_2{\mathrm{H}}_2}\left(\mathrm{g}\right)+\underset{\frac{5}{2}\mathrm{z} \mathrm{ml}}{\frac{5}{2}{\mathrm{O}}_2}\left(\mathrm{g}\right) \rightleftharpoons 2{\mathrm{CO}}_2\left(\mathrm{g}\right)+{\mathrm{H}}_2\mathrm{O}\left(\mathcal{l}\right) \end{gathered}$$

        Total volume of oxygen used up in the reaction $=\left(2\mathrm{x}+3\mathrm{y}+\frac{5}{2}\mathrm{z}\right)\mathrm{ml}$

        But from question,

        Total volume of oxygen used up $=30-5.5=24.5 \mathrm{ml}$

$\therefore 2\mathrm{x}+3\mathrm{y}+\frac{5}{2}\mathrm{z}=24.5$                 ................... (3)

         Solving equations (1), (2) and (3), we get

          x=4, y=3, z=3

$\therefore {\mathrm{CH}}_4=4\mathrm{ml}, {\mathrm{C}}_2{\mathrm{H}}_4=3\mathrm{ml}, {\mathrm{C}}_2{\mathrm{H}}_2=3 \mathrm{ml}$

$$\begin{gathered} \left(\mathrm{A}\right). 200 \min \mathrm{pressure} \mathrm{of} {\mathrm{O}}_2\stackrel{ \mathrm{t}=47 \min }{\to }1500 \mathrm{mm} \\ 4000 \mathrm{mm} \mathrm{pressure} \mathrm{of} \mathrm{mixture} \stackrel{ \mathrm{t}=74 \min }{\to }1 : 1 \left({\mathrm{O}}_2+\mathrm{gas}\right) \\ \mathrm{For} \mathrm{pure} {\mathrm{O}}_2, \frac{{\mathrm{P}}_1}{{\mathrm{P}}_2}=\frac{{\mathrm{n}}_1}{{\mathrm{n}}_2} \\ \mathrm{Where} {\mathrm{n}}_1 \mathrm{and} {\mathrm{n}}_2 \mathrm{are} \mathrm{original} \mathrm{no}. \mathrm{of} \mathrm{moles} {\mathrm{O}}_2 \mathrm{and} \mathrm{moles} \mathrm{of} {\mathrm{O}}_2 \mathrm{after} 47 \mathrm{minutes}. \\ \therefore \frac{{\mathrm{n}}_1}{{\mathrm{n}}_2}=\frac{2000}{1500} \therefore {\mathrm{n}}_2=\frac{3}{4{\mathrm{n}}_1} \\ \mathrm{or} \mathrm{moles} \mathrm{of} {\mathrm{O}}_2 \mathrm{diffused} \mathrm{in} 47 \mathrm{minutes}={\mathrm{n}}_1-\frac{3{\mathrm{n}}_1}{4}=\frac{{\mathrm{n}}_1}{4} \\ \mathrm{or} \mathrm{moles} \mathrm{of} {\mathrm{O}}_2 \mathrm{diffused} \mathrm{in} 74 \mathrm{minutes}=\frac{{\mathrm{n}}_1\times 74}{47\times 4} \\ =\frac{74}{188}{\mathrm{n}}_1=\frac{74}{188}\left(\mathrm{if} {\mathrm{n}}_1=1\right)=0.3936 \\ \mathrm{Since}, \mathrm{diffusion} \mathrm{of} {\mathrm{O}}_2 \mathrm{in} \mathrm{mixture} \mathrm{also} \mathrm{occurs} \mathrm{at} \mathrm{partial} \mathrm{pressure} \mathrm{of} 200 \mathrm{mm} \left(\mathrm{The} \mathrm{ratio} \mathrm{of} \mathrm{gas} \mathrm{and} {\mathrm{O}}_2 \mathrm{being} 1:1\right) \\ \mathrm{Now}, \mathrm{gas} \mathrm{and} {\mathrm{O}}_2 \mathrm{both} \mathrm{diffusing} \mathrm{in} \mathrm{the} \mathrm{form} \mathrm{of} \mathrm{mixture} \mathrm{through} \mathrm{same} \mathrm{orifice} \mathrm{at} \mathrm{partial} \mathrm{pressure} \mathrm{of} 2000 \mathrm{mm} \mathrm{each}. \\ \therefore \frac{{\mathrm{n}}_{{\mathbf{O}}_{\mathbf{2}}}}{74}\times \frac{74}{{\mathrm{n}}_{\mathrm{g}}}=\sqrt{\left(\frac{79}{32}\right)} \\ \therefore {\mathrm{n}}_{\mathrm{g}}={\mathrm{n}}_{{\mathbf{O}}_{\mathbf{2}}}\times \sqrt{\left(\frac{32}{79}\right)}=\frac{74}{188}\times \sqrt{\left(\frac{32}{79}\right)}=0.249 \\ \therefore \mathrm{Moles} \mathrm{of} {\mathrm{O}}_2 \mathrm{left} \mathrm{after} 74 \mathrm{minutes} = 1-0.3936=0.06064 \\ \mathrm{Moles} \mathrm{of} \mathrm{gas} \mathrm{after} 74 \mathrm{minutes}=1-0.249=0.7510 \\ \therefore {\mathrm{O}}_2 : \mathrm{gas}=0.6064 : 0.7510 \\ \therefore 1 : 1.236 \end{gathered}$$

1)

when temp became twice, vol will become twice ( Boyle's law)

PV = nRT , n =  moles of water vapour

$n=\frac{PV}{RT}=\frac{24.8\times 2}{0.0821\times 393}=1.53 moles$

wt of ${\mathrm{MgSO}}_4$ = 54-27.6 = 26.5

4.

Let x be the mass of Ne and y be the mass of Ar

x + y = 30                                 ...(i)

The molar mass of Ar = 40 g

The molar mass of Ne = 20 g

Moles of Neon + moles of Argon = Total noles

$$\begin{gathered} \frac{\mathrm{x}}{20} + \frac{\mathrm{y}}{40} = \mathrm{n} \\ \mathrm{Total} \mathrm{of} \mathrm{K}.\mathrm{E}. = \frac{3}{2}\mathrm{nRT} \\ \mathrm{Given}, \\ \mathrm{K}.\mathrm{E}. = 3 \mathrm{KJ}, \mathrm{T} = 250 \mathrm{K}, \mathrm{R} = 8.314 \mathrm{J} {\mathrm{K}}^{-1} {\mathrm{mol}}^{-1} \\ 3000 \mathrm{J} = \frac{3}{2}\left(\frac{\mathrm{x}}{20} + \frac{\mathrm{y}}{40}\right) \times 8.314 \times 250 \\ 2\mathrm{x} + \mathrm{y} = 38.48 ...\left(\mathrm{ii}\right) \\ \mathrm{From} \left(\mathrm{i}\right) \mathrm{and} \left(\mathrm{ii}\right) \\ \mathrm{y} = 21.52 \mathrm{g} \\ \mathrm{mass} \mathrm{of} \mathrm{Neon} = \frac{\mathrm{Massof} \mathrm{neon}}{\mathrm{Total} \mathrm{mass}} \times 100 \\ = \frac{8.48}{30} \times 100 = 28.3\% \end{gathered}$$

K.E = 3/2 RT, so does not depend on moles