JEE Advanced 2016 Paper 1 Question Paper with Solutions

The radius of $n^{\text {th }}$ orbital for a hydrogen-like atom of atomic number $Z$ is given by $r_n={{{n^2}{a_0}} \over Z}$, where a_0=0.53 <span class="katex-error" title="ParseError: KaTeX parse error: Can&#x27;t use function &#x27;' in math mode at position 1: ̲\mathop A\limit…" style="color:#cc0000">\mathop A\limits^o is the Bohr's radius.

$\therefore$ $\text { Radius of orbit, } r \propto \frac{n^2}{Z}$

The relative change in the radii of two consecutive orbitals is

$\frac{r_{n+1}-r_n}{r_n}=\frac{\frac{(n+1)^2}{Z}-\frac{n^2}{Z}}{\frac{n^2}{Z}}=\frac{2 n+1}{n^2} \approx \frac{2}{n} \quad(\because n>>1)$

The energy of the n^th orbital is given by

${E_n} = {{ - 13.6{Z^2}} \over {{n^2}}}eV$

$\therefore$ $E_n \propto \frac{-Z^2}{n^2}$

The relative change in the energy of two consecutive orbitals is

$$\begin{aligned} & \frac{E_{n+1}-E_n}{E_n}=\frac{\frac{-Z^2}{(n+1)^2}+\frac{Z^2}{n^2}}{\frac{-Z^2}{n^2}}=-\frac{(2 n+1)}{(n+1)^2} \\ & \approx-\frac{2 n}{n^2}=-\frac{2}{n} \quad(n>>1) \end{aligned}$$

The angular momentum of the $n^{\text {th }}$ orbital is given by $L_n={{nh} \over {2\pi }}$. The relative change in the angular momentum of two consecutive orbitals is

$\frac{L_{n+1}-L_n}{L_n}=\frac{(n+1) {{h} \over {2\pi }}-{{nh} \over {2\pi }}}{{nh} \over {2\pi }}=\frac{1}{n}$

${{hc} \over \lambda } - \phi = e{V_0}$ ($\phi$ = work function)

${{hc} \over {0.3 \times {{10}^{ - 6}}}} - \phi = 2e$ .... (i)

${{hc} \over {0.4 \times {{10}^{ - 6}}}} - \phi = 1e$ .... (ii)

Subtracting Eq. (ii) from Eq. (i)

$hc\left( {{1 \over {0.3}} - {1 \over {0.4}}} \right){10^6} = e$

$hc\left( {{{0.1} \over {0.12}} \times {{10}^6}} \right) = e$

$h = 0.64 \times {10^{ - 33}}$

$= 6.4 \times {10^{ - 34}}$ J-s

${\log _2} = {{{p_1}} \over {{p_0}}} = 1$

Therefore, ${{{p_1}} \over {{p_0}}} = 2$

According to Stefan's law, p $\propto$ T²

$\Rightarrow {{{p_2}} \over {{p_1}}} = {\left( {{{{T_2}} \over {{T_1}}}} \right)^4} = {\left( {{{2767 + 273} \over {487 + 273}}} \right)^4} = {4^4}$

${{{p_2}} \over {{p_1}}} = {{{p_2}} \over {2{p_0}}} = {4^4} \Rightarrow {{{p_2}} \over {{p_0}}} = 2 \times {4^4}$

${\log _2}{{{p_2}} \over {{p_0}}} = {\log _2}[2 \times {4^4}]$

$= {\log _2}2 + {\log _2}{4^4}$

$= 1 + {\log _2}{2^8} = 1 + 8 = 9$

For refraction through plano - Convex lens,

$m=-2=\frac{v}{u}$

$$\begin{array}{ll} \Rightarrow v=-2 u \\ \text { As, } u=-30 \mathrm{~cm} \\ \Rightarrow v=60 \mathrm{~cm} \end{array}$$

Using lens formula, $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$

$$\begin{aligned} \frac{1}{60}-\frac{1}{-30} & =\frac{1}{f} \\ \Rightarrow f =20 \mathrm{~cm} \end{aligned}$$



By lens maker's formula,

$$\begin{aligned} \frac{1}{f} & =(n-1)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right) \\ \Rightarrow \frac{1}{20} & =(n-1)\left(\frac{1}{\mathrm{R}}+\frac{1}{\infty}\right)=\frac{n-1}{\mathrm{R}} \\ \Rightarrow n & =1+\left(\frac{\mathrm{R}}{20}\right) ........(i) \end{aligned}$$

for poor reflection from convex surface,

$$\begin{aligned} & u=-30 \mathrm{~cm}, \\ & v=10 \mathrm{~cm}, f=\mathrm{R} / 2 \end{aligned}$$



Using mirror formula,

$$\begin{aligned} \frac{1}{v}+\frac{1}{u} & =\frac{1}{f}=\frac{2}{\mathrm{R}} ; \\ \Rightarrow \frac{1}{10}+\frac{1}{-30} & =\frac{2}{\mathrm{R}} \end{aligned}$$

$$\begin{array}{llrl} \Rightarrow \frac{2}{\mathrm{R}} =\frac{2}{30} \\ \therefore \mathrm{R} =30 \mathrm{~cm} \end{array}$$

Substitute $R=30 \mathrm{~cm}$ in equation (i) to get $n=$ 2.5 for $f=20 \mathrm{~cm}$ (inverted image) and $n=1.5$ for $f=60 \mathrm{~cm}$ (erect image).

The energy of the incident photon of wavelength $\lambda=970$ $\mathop A\limits^o$ is

$\Delta E=\frac{h c}{\lambda}=\frac{1.237 \times 10^{-6}}{970 \times 10^{-10}}=12.75 \mathrm{eV}$

$$\begin{aligned} & \text { Final energy of electron in } n^{\text {th }} \text { state }=-\frac{13.6}{n^2} \mathrm{eV} \\ & \Rightarrow-\frac{13.6}{n^2}=-13.6+12.75=-0.85 \\ & \Rightarrow n^2=16 \Rightarrow n=4 \end{aligned}$$



The hydrogen atom can make ${ }^n \mathrm{C}_2=6$ transitions while returning to ground state. Thus, the emission spectrum will have six lines.

The circuit is shown in the figure

Initially, right after the circuit is switched on ($t \rightarrow 0^{+}$), the impedance (effective resistance) of inductors $L_1$ and $L_2$ is extremely high. Therefore, the inductors act as open circuits, and all the current flows through the resistor $R = 12 \Omega$. According to Kirchhoff's law, the current through the battery at this moment is $i_{\min} = \frac{V}{R} = \frac{5}{12} \, \text{A}$.

As the circuit reaches steady state ($t \rightarrow \infty$), the impedance of the inductors drops to zero, and they function as resistors with their specified internal resistances.

The effective resistance of the circuit is $R_e=(12 \Omega \|$ $4 \Omega)\|3 \Omega=3 \Omega\| 3 \Omega=3 / 2 \Omega$.

In the steady state, the current through the circuit is $i_{\max} = \frac{V}{R_e} = \frac{5}{\frac{3}{2}} = \frac{10}{3} \, \text{A}$.

Therefore, the ratio of the maximum current to the minimum current is:

$\frac{i_{\max}}{i_{\min}} = \frac{\frac{10}{3}}{\frac{5}{12}} = \frac{10}{3} \times \frac{12}{5} = 8$

Heat generated in device in 3 h

= Time $\times$ power

= 3 $\times$ 3600 $\times$ 3 $\times$ 10³ = 324 $\times$ 10⁵ J

Heat used to heat water = ms $\Delta$ $\theta$

= 120 $\times$ 1 $\times$ 4.2 $\times$ 10³ $\times$ 20J

Heat absorbed by coolant

Pt = 324 $\times$ 10⁵ $-$ 120 $\times$ 1 $\times$ 4.2 $\times$ 10³ $\times$ 20J

Pt = (325 $-$ 100.8) $\times$ 10⁵ J = 223.2 $\times$ 10⁵ J

P = ${{223.2 \times {{10}^5}} \over {3600}}$ = 2067 W

$\sum {{F_x} = 0}$, ${N_1}\cos 30^\circ - f = 0$ .... (i)

$\sum {{F_y} = 0}$, ${N_1}\sin 30^\circ + {N_2} - mg = 0$ .... (ii)

$\sum {{\tau _0} = 0}$

$mg{1 \over 2}\cos 60^\circ - {N_1}{h \over {\cos 30^\circ }} = 0$ .... (iii)

Also, given ${N_1} = {N_2}$ .... (iv)

Solving Eqs. (i), (ii), (iii) and (iv) we have

${h \over l} = {{3\sqrt 3 } \over {16}}$ and $f = {{16\sqrt 3 } \over 3}$

Applying Snell's law at M,

$n = {{\sin \alpha } \over {\sin {r_1}}} \Rightarrow \sqrt 2 = {{\sin 45^\circ } \over {\sin {r_1}}}$

$\Rightarrow \sin {r_1} = {{\sin 45^\circ } \over {\sqrt 2 }} = {{1/\sqrt 2 } \over {\sqrt 2 }} = {1 \over 2}$

${r_1} = 30^\circ$

$\sin {\theta _c} = {1 \over n} = {1 \over {\sqrt 2 }} \Rightarrow {\theta _c} = 45^\circ$

Let us take ${r_2} = {\theta _c} = 45^\circ$ for just satisfying the condition of TIR.

In $\Delta PNM$,

$\theta + 90 + {r_1} + 90 - {r_2} = 180^\circ$

or $\theta = {r_2} - {r_1} = 45^\circ - 30^\circ = 15^\circ$

Note If $\alpha$ = 45$^\circ$ (the given value). Then, r₁ > 30$^\circ$ (the obtained value)

${r_2} > {\theta _c}$ (as r₂ $-$ r₁ = $\theta$ or r₂ = $\theta$ + r₁)

or TIR will take place. So, for taking TIR under all conditions $\alpha$ should be greater than 45$^\circ$ or this is the minimum value of $\alpha$.

The flux passing through the triangular wire if ' $i$ ' current flows through the infinitely long conducting wire.


$$\begin{aligned} \frac{d \mathrm{I}}{d t} & =10 \mathrm{As}^{-1} \\ \phi & =\int_0^{0.1} \frac{\mu_0 i}{2 \pi x} \times 2 x d x \\ \phi & =\frac{\mu_0 i}{10 \pi}=\mathrm{M} i \end{aligned}$$

$\therefore \mathrm{M}=\frac{\mu_0}{10 \pi}$

$\text { Induced emf in the wire }=\frac{\mathrm{M} d i}{d t}=\frac{\mu_0}{10 \pi} \times 10=\frac{\mu_0}{\pi}$

According to Lenz's law, the induced current opposes the flux change that creates it. Since the flux is increasing, the induced emf opposes the existing magnetic field and the current flows from left to right in the straight wire, that is, along the hypotenuse current. That is, there is a repulsive force between the wire and the loop.


Also, if the loop is rotated about the wire, there is no change in the flux through the wire, so no extra induced emf is generated in the wire.

$r = \alpha {t^3}\widehat i + \beta {t^2}\widehat j$

$v = {{dr} \over {dt}} = 3\alpha {t^2}\widehat i + 2\beta {t^2}\widehat j$

$a = {{{d^2}r} \over {d{t^2}}} = 6\alpha {t^2}\widehat i + 2\beta {t^2}\widehat j$

At t = 1 s,

(a) $v = 3 \times {{10} \over 3} \times 1\widehat i + 2 \times 5 \times 1\widehat j$

$= (10\widehat i + 10\widehat j)$ m/s

(b) $\widehat L = \widehat r \times \widehat p$

$= \left( {{{10} \over 3} \times 1\widehat i + 5 \times 1\widehat j} \right) \times 0.1(10\widehat i + 10\widehat j)$

$= \left( { - {5 \over 3}\widehat k} \right)$ N-ms

(c) $F = ma$

$= m \times \left( {6 \times {{10} \over 3} \times 1\widehat i + 2 \times 5\widehat j} \right) = (2\widehat i + \widehat j)N$

(d) $\tau = r \times F = \left( {{{10} \over 3}\widehat i + 5\widehat j} \right) \times (2\widehat i + \widehat j)$

$= + {{10} \over 3}\widehat k + 10( - \widehat k) = \left( { - {{20} \over 3}\widehat k} \right)$ N-m

From Snell's law,

$$\begin{aligned} n_1 \sin \theta_i=n\left(z_1\right) \sin \theta_{z_1}=n\left(z_2\right) \sin \theta_{z_2} & =\ldots \ldots \\ \ldots & =n(d) \sin \theta_d=n_2 \sin \theta_f \end{aligned}$$

The lateral displacement of a ray by a parallel slab depends on the angle of incidence $\theta_i$, refractive index $n_1$, slab thickness $d$ and the refractive index of the slab $n$ (it is independent of $n_2$ because $l$ is measured before the start of medium 2, there is no effect of $n_2$ on $l$ ). Consider the parallel slab shown in the figure.


By Snell's law, $n_1 \sin i_1=n \sin r_1$ and $n \sin i_2=$ $n_2 \sin r_2$. By geometry, $i_2=r_1$ and the lateral displacement is given by

$l=d \frac{\sin r_1}{\cos r_1}=\frac{d n_1 \sin i_1}{\sqrt{n^2-n_1^2 \sin ^2 i_1}} .$

Hence, $l$ is independent of $n_2$ but it is dependent of $n(z)$.

Consider the time $t\left(\leq t_Q\right)$ when car is at $S_1$ between $P$ and $Q$. The distance travelled by the car in time $t$ is $\mathrm{PS}_1=u t$. At this instant, the lines $\mathrm{S}_1{N}$ and $\mathrm{S}_1 \mathrm{M}$ both make angle $\theta$ with the velocity vector $\vec{u}$. The component of observer (person sitting in the car) velocity towards the sources $N$ and $M$ is $u_o=u \cos \theta$. The sources $N$ and $M$ are at rest i.e., $u_s=0$. Apply Doppler's effect equation to get frequencies of the sources $N$ and $M$ heard by the observer as

$$\begin{aligned} \nu_N^{\prime} & =\frac{v+u_o}{v-u_s} \nu_N=\frac{v+u \cos \theta}{v} \nu_N \\ & =\left(1+\frac{u}{v} \frac{D-u t}{\sqrt{d^2+(D-u t)^2}}\right) \nu_N, \\ \nu_M^{\prime} & =\frac{v+u_o}{v-u_s} \nu_M=\frac{v+u \cos \theta}{v} \nu_M \\ & =\left(1+\frac{u}{v} \frac{D-u t}{\sqrt{d^2+(D-u t)^2}}\right) \nu_M . \end{aligned}$$

The beat frequency heard by the observer at time $t(\leq$ $\left.t_Q\right)$ is

$$\begin{aligned} \nu(t) & =\nu_N^{\prime}-\nu_M^{\prime} \\ & =\left(1+\frac{u}{v} \frac{D-u t}{\sqrt{d^2+(D-u t)^2}}\right)\left(\nu_N-\nu_M\right) ........(1) \end{aligned}$$

Now, consider the time $t\left(\geq t_Q\right)$ when car is at $S_2$ between $Q$ and $R$. The distance travelled by the car in time $t$ is $\mathrm{PS}_2=u t$. At this instant, the lines $\mathrm{S}_2 \mathrm{~N}$ and $\mathrm{S}_2 \mathrm{M}$ both make angle $\left(180^{\circ}-\theta\right)$ with the velocity vector $\vec{u}$. The component of observer velocity towards the sources $N$ and $M$ is $u_o=-u \cos \theta$. Apply Doppler's effect equation to get

$$\begin{aligned} \nu_N^{\prime} & =\frac{v+u_o}{v-u_s} \nu_N=\frac{v-u \cos \theta}{v} \nu_N \\ & =\left(1-\frac{u}{v} \frac{u t-D}{\sqrt{d^2+(u t-D)^2}}\right) \nu_N, \\ \nu_M^{\prime} & =\frac{v+u_o}{v-u_s} \nu_M=\frac{v-u \cos \theta}{v} \nu_M \\ & =\left(1-\frac{u}{v} \frac{u t-D}{\sqrt{d^2+(u t-D)^2}}\right) \nu_M . \end{aligned}$$

The beat frequency heard by the observer at time $t(\geq$ $\left.t_Q\right)$ is

$$\begin{aligned} \nu(t) & =\nu_N^{\prime}-\nu_M^{\prime} \\ & =\left(1-\frac{u}{v} \frac{u t-D}{\sqrt{d^2+(u t-D)^2}}\right)\left(\nu_N-\nu_M\right) ..........(2) \end{aligned}$$

Substitute $t=0$ and $t=t_Q={D \over u}$ in equation (1) to get $\nu_P$ and $\nu_Q$ and substitute $t=t_R=2 {D \over u}$ in equation (2) to get $\nu_R$ i.e.,

$$\begin{aligned} & \nu_P=\nu(t=0)=\left(1+\frac{u}{v} \frac{D}{\sqrt{d^2+D^2}}\right)\left(\nu_N-\nu_M\right) \\ & \nu_Q=\nu(t=D / u)=\left(\nu_N-\nu_M\right) \\ & \nu_R=\nu(t=2 D / u)=\left(1-\frac{u}{v} \frac{D}{\sqrt{d^2+D^2}}\right)\left(\nu_N-\nu_M\right) \end{aligned}$$

$\therefore$ $\nu_P+\nu_R=2 \nu_Q$

Differentiate equations (1) w.r.t. time t to get rate of change of beat frequency

$\frac{\mathrm{d} \nu(t)}{\mathrm{d} t}=-\left(\nu_N-\nu_M\right) \frac{u^2}{v} \frac{d^2}{\left(d^2+(D-u t)^2\right)^{3 / 2}}$ ...........(3)

From equation (3), the slope is negative and its magnitude is maximum when $t={D \over u}=t_Q$ (denominator is minimum). Thus the rate of change of beat frequency is maximum when car passes through $Q$.

Terminal velocity is given by

${v_T} = {2 \over 9}{{{r^2}} \over \eta }(d - \rho )g$

${{{v_P}} \over {{v_Q}}} = {{r_P^2} \over {r_Q^2}} \times {{{\eta _Q}} \over {{\eta _P}}} \times {{(d - {\rho _P})} \over {(d - {\rho _Q})}}$

$= {\left( {{1 \over {0.5}}} \right)^2} \times \left( {{2 \over 3}} \right) \times {{(8 - 08)} \over {(8 - 1.6)}} = 4 \times {2 \over 3} \times {{7.2} \over {6.4}} = 3$

Because of non-uniform evaporation at different section, area of cross-section would be different at different sections.

Region of highest evaporation rate would have rapidly reduced area and would become break up cross-section.

Resistance of the wire as whole increases with time. Overall resistance increases hence power decreases.

($p = {{{V^2}} \over R}$ or $p \propto {1 \over R}$ as V is constant).

At break up junction temperature would be highest, thus light of highest band frequency would be emitted at those cross-section.

Let the charge per unit length on the axis be $\lambda$(t) at a time t. The electric field due to this line charge at a radial distance r is given by

$\overrightarrow E (r,t) = {{\lambda (t)} \over {2\pi \in r}}\widehat r$. ...... (1)

The free charges inside the conductor start moving radially due to the presence of electric field. The current density at a point is defined as the current flowing across a unit area placed perpendicular to the direction of current flow. The current density at a point is related to the electric field at that point by Ohm's law i.e.,

$\overrightarrow j (r,t) = \sigma \overrightarrow E (r,t)$, ....... (2)

where $\sigma$ is the electrical conductivity of the conductor. The current through a cylindrical shell of radius r and length l is given by

$I = \int_{surface} {\overrightarrow j (r,t)\,.\,d\overrightarrow A = j(r,t)(2\pi rl)}$. ...... (3)

By conservation of charge, the current I is equal to the rate of decrease of charge q on axial line segment of length l i.e.,

$I = - {{dq} \over {dt}} = - l{{d\lambda (t)} \over {dt}}$. ...... (4)

From equations (1) - (4)

${{d\lambda (t)} \over {dt}} = - {{\sigma \lambda (t)} \over \in }$.

Integrate with initial condition $\lambda$(t = 0) = $\lambda$₀ to get

$\lambda (t) = {\lambda _0}\exp {e^{ - {{\sigma t} \over \in }}}$,

Substitute $\lambda$(t) in equation (2) to get

$\overrightarrow j (r,t) = {{\sigma {\lambda _0}} \over {2\pi \in r}}{e^{ - {{\sigma t} \over \in }}}$.

Note that current density varies as 1/r with radial distance r. It decreases exponentially with time and becomes zero at t $\to$ $\infty$.

As the ${ }_5^{12} B$ is going under $\beta$-decay to form $_6^{12}C$. We can write the balanced equation as follows

${ }_5^{12} \mathrm{~B} \rightarrow{ }_6^{12} \mathrm{C}+e^{-1}+\bar{\nu}$

The $Q$-value of this reaction is given by

$$\begin{aligned} Q & =\left[m\left({ }_5^{12} \mathrm{~B}\right)-m\left({ }_6^{12} \mathrm{C}\right)\right] c^2 \\ & =[12.041-12.0] \times 931.5=13.041 \mathrm{MeV} \end{aligned}$$

The energy $Q=13.041 \mathrm{MeV}$ is released in the reaction. Out of this energy, 4.041 $\mathrm{MeV}$ is used to excite ${ }_6^{12} \mathrm{C}$ to its excited state ${ }_6^{12} \mathrm{C}^*$. Thus, the kinetic energy available to the $\beta$-particle $\left(K_\beta\right)$ and the antineutrino $\left(K_{\bar{\nu}}\right)$ is $K_\beta+K_{\bar{\nu}}=13.041-4.041=9 \mathrm{MeV}$. In $\beta-$ decay, the kinetic energy of the $\bar{\nu}$ can vary from zero to a maximum value. Hence, the maximum kinetic energy of the $\beta$-particle is $K_{\beta, \max }=9 \mathrm{MeV}$ (when $K_{\bar{\nu}}=0$ ).

$[n] = [{L^{ - 3}}];[q] = [AT]$

$[\varepsilon ] = [{M^{ - 1}}{L^{ - 3}}{A^2}{T^4}]$

$[T] = [L]$

$[l] = [L]$

$[{k_B}] = [{M^1}{L^2}{T^{ - 2}}{K^{ - 1}}]$

(a) RHS

$= \sqrt {{{[{L^{ - 3}}{A^2}{T^2}]} \over {[{M^{ - 1}}{L^{ - 3}}{T^4}{A^2}][{M^1}{L^2}{T^{ - 2}}{K^{ - 1}}][K]}}}$

$= \sqrt {{{[{L^{ - 3}}{A^2}{T^2}]} \over {[{L^{ - 1}}{A^2}{T^2}]}}} = \sqrt {[{L^{ - 2}}]} = [{L^{ - 1}}]$ Wrong

(b) RHS

$= \sqrt {{{[{M^{ - 1}}{L^{ - 3}}{T^4}{A^2}][{M^1}{L^2}{T^{ - 2}}{K^{ - 1}}][K]} \over {[{L^{ - 3}}{A^2}{T^2}]}}}$

$= \sqrt {{{[{L^{ - 1}}{A^2}{T^2}]} \over {[{L^{ - 3}}{A^2}{T^2}]}}} = L$ Correct

(c) RHS

$= \sqrt {{{[{A^2}{T^2}]} \over {[{M^{ - 1}}{L^{ - 3}}{T^4}{A^2}][{L^{ - 2}}][{M^1}{L^2}{T^{ - 2}}{K^{ - 1}}][K]}}}$

$= \sqrt {[{L^3}]}$ Wrong

(d) RHS

$= \sqrt {{{[{A^2}{T^2}]} \over {[{M^{ - 1}}{L^{ - 3}}{T^4}{A^2}][{L^{ - 1}}][{M^1}{L^2}{T^{ - 2}}{K^{ - 1}}][K]}}}$

$= \sqrt {{{[{A^2}{T^2}]} \over {[{L^{ - 2}}{T^2}{A^2}]}}} = [L]$ Correct

Structure of crystalline form of borax


(a) Borax contains four boron atom; hence, it is tetranuclear.

(b) Only 2 boron atoms lie in the same plane.

(c) Two boron atoms are $s p^3$ hybridised other two borons are $s p^2$ hybridised.

(d) Each of the boron has hydroxyl group attached to it with two hydroxyl groups present as hydroxyl salts.

It is given that the mole fraction of the solute $\left(x_{\text {solute }}\right)$ is 0.1 , therefore mole fraction of the solvent $\left(x_{\text {solvent }}\right)$ is 0.9 . So we have

$x_{\text {solute }}=\frac{n_{\text {solute }}}{\left(n_{\text {solute }}+n_{\text {solvent }}\right)}=0.1$ ......(1)

$x_{\text {solvent }}=\frac{n_{\text {solvent }}}{\left(n_{\text {solute }}+n_{\text {solvent }}\right)}=0.9$ ......(2)

where $n_{\text {solute }}$ and $n_{\text {solvent }}$ are the number of moles of solute and solvent, respectively. Dividing Eq. (1) by Eq. (2) gives

$\frac{x_{\text {solute }}}{x_{\text {solvent }}}=\frac{n_{\text {solute }}}{n_{\text {solvent }}}=\frac{0.1}{0.9}$ .......(3)

Given that the density of solution is $2 \mathrm{~g} \mathrm{~cm}^{-3}$, we have

$$\begin{aligned} & W_{\text {solution }} =\text { density } \times V_{\text {solution }}=2 \times V_{\text {solution }} \\ & \therefore W_{\text {solute }}+W_{\text {solvent }} =2 \times V_{\text {solution }} .......(4) \end{aligned}$$

We know that molality is given by

$m=\frac{n_{\text {solute }} \times 1000}{n_{\text {solvent }} \times W_{\text {solvent }}}$

Substituting from Eq. (3), we have

$m=\left(\frac{0.1}{0.9}\right) \times \frac{1000}{W_{\text {solvent }}}$ ......(5)

Molarity is given by

$M=\frac{n_{\text {solute }} \times 1000}{n_{\text {solvent }} \times V_{\text {solution }}}$

Substituting from Eq. (3), we have

$M=\left(\frac{0.1}{0.9}\right) \times \frac{1000}{V_{\text {solution }}}$ ..........(6)

Given that, Molarity $(M)=$ Molality $(m)$;

Therefore, from Eqs. (5) and (6), we get

$$\begin{aligned} & \frac{0.1 \times 1000}{0.9 \times W_{\text {solvent }}} =\frac{0.1 \times 1000}{0.9 \times V_{\text {solution }}} \\ & W_{\text {solvent }} =V_{\text {solution }} \end{aligned}$$

From Eq. (4), we have

$$\begin{aligned} W_{\text {solvent }} & =\frac{W_{\text {solute }}+W_{\text {solvent }}}{2} \\ 2 W_{\text {solvent }} & =W_{\text {solute }}+W_{\text {solvent }} \\ W_{\text {solvent }} & =W_{\text {solute }} .........(7) \end{aligned}$$

The molecular weight (MW) of the solute can be calculated by dividing the weight of the solute by the number of moles of solute. This can be written as :

$MW_{\text{solute}} = \frac{W_{\text{solute}}}{n_{\text{solute}}}$

2. Similarly, the molecular weight (MW) of the solvent can be calculated by dividing the weight of the solvent by the number of moles of solvent. This can be written as :

$MW_{\text{solvent}} = \frac{W_{\text{solvent}}}{n_{\text{solvent}}}$

In these formulas, $MW_{\text{solute}}$ and $MW_{\text{solvent}}$ represent the molecular weights of the solute and solvent respectively, $W_{\text{solute}}$ and $W_{\text{solvent}}$ represent their weights, and $n_{\text{solute}}$ and $n_{\text{solvent}}$ represent the number of moles of solute and solvent respectively.

From Eq. (3), we have

$\frac{n_{\text {solute }}}{n_{\text {solvent }}}=\frac{W_{\text {solute }} / M W_{\text {solute }}}{W_{\text {solvent }} / M W_{\text {solvent }}}=\frac{0.1}{0.9}$

Using Eq. (7), we get

$\frac{M W_{\text {solute }}}{M W_{\text {solvent }}}=9$

Total five isomers are possible :

Consider the reactions of the given compounds with $\mathrm{S}^{2-}$ and $\mathrm{SO}_4^{2-}$.

(i) With $\mathrm{CuCl}_2$ :

$$\begin{aligned} &\mathrm{CuCl}_2+\mathrm{S}^{2-} \rightarrow \underset{\text { black ppt }}{\mathrm{CuS}}+2 \mathrm{Cl}^{-} \\ &\mathrm{CuCl}_2+\mathrm{SO}_4{ }^{2-} \rightarrow \text { No precipitate } \end{aligned}$$

(ii) With $\mathrm{BaCl}_2$ : Alkaline earth metal sulphides are sparingly soluble in water but pass into solution with time and hence do not give precipitates from their aqueous solutions.

$$\begin{aligned} &\mathrm{BaCl}_2+\mathrm{S}^{2-} \rightarrow \underset{\text { No ppt }}{\mathrm{BaS}}+2 \mathrm{Cl}^{-} \\ &\mathrm{BaS}+2 \mathrm{H}_2 \mathrm{O} \rightleftharpoons \underset{\text { water soluble }}{\mathrm{Ba}(\mathrm{OH})_2+{\mathrm{Ba}(\mathrm{SH})_2}} \end{aligned}$$

Alkaline earth metal sulphates, $\mathrm{BaSO}_4, \mathrm{CaSO}_4$ and $\mathrm{SrSO}_4$ are insoluble in water.

$\mathrm{BaCl}_2+\mathrm{SO}_4^{2-} \rightarrow \underset{\text { white ppt }}{\mathrm{BaSO}_4}+2 \mathrm{Cl}^{-}$

(iii) With $\mathrm{Pb}(\mathrm{OAc})_2$ :

$$\begin{aligned} &\mathrm{Pb}(\mathrm{OAc})_2+\mathrm{S}^{2-} \rightarrow \underset{\text { black ppt }}{\mathrm{PbS}}+2 \mathrm{CH}_3 \mathrm{COO}^{-} \\ &\mathrm{Pb}(\mathrm{OAc})_2+\mathrm{SO}_4^{2-} \rightarrow \underset{\text { white ppt }}{\mathrm{PbSO}_4}+2 \mathrm{CH}_3 \mathrm{COO}^{-} \end{aligned}$$

The $K_{\mathrm{sp}}$ of $\mathrm{PbS}$ is $3 \times 10^{-28}$ and of $\mathrm{PbSO}_4$ is $25 \times$ $10^{-28}$. The difference is large, thus only $\mathrm{PbS}$ is selectively precipitated.

(iv) With $\mathrm{Na}_2\left[\mathrm{Fe}(\mathrm{CN})_5 \mathrm{NO}\right]$ : When sodium nitroprusside is added to solution of sulphide ions, purple colouration is observed. This is a confirmative test for sulphides.

$$\begin{aligned} &{\left[\mathrm{Fe}(\mathrm{CN})_5 \mathrm{NO}\right]^{2-}+\mathrm{S}^{2-} \rightarrow \underset{\text { Purple colour }}{\left[\mathrm{Fe}(\mathrm{CN})_5 \mathrm{NOS}\right]}} \\ &\mathrm{Na}_2\left[\mathrm{Fe}(\mathrm{CN})_5 \mathrm{NO}\right]+\mathrm{SO}_4{ }^{2-} \rightarrow \text { No precipitate } \end{aligned}$$