MHT CET 2020 Chemistry Question Paper with Answer and Solution

(C) Dacron is manufactured by heating a mixture of ethylene glycol and terephthalic acid at $420-460 \ K$ in the presence of zinc acetate-antimony trioxide catalyst.

(C) Neoprene is a synthetic rubber formed by the free radical polymerization of chloroprene ($2$-chlorobuta-$1,3$-diene).
The chemical reaction is as follows:
$n \ CH_2=C(Cl)-CH=CH_2 \xrightarrow{\text{polymerization}} [-CH_2-C(Cl)=CH-CH_2-]_n$
Thus,chloroprene is the monomer used for the preparation of neoprene.

(A) The oxidation state of $K$ is $+1$ and the oxalate ion $(C_{2}O_{4}^{2-})$ has a charge of $-2$.
Let the oxidation state of $Cr$ be $x$.
For the complex $K_{3}[Cr(C_{2}O_{4})_{3}]$,the sum of oxidation states is zero:
$(3 \times (+1)) + x + (3 \times (-2)) = 0$
$3 + x - 6 = 0$
$x - 3 = 0$
$x = +3$
Therefore,the oxidation state of $Cr$ is $+3$.

(D) Let the oxidation number of $Ru$ be $x$.
$NH_{3}$ is a neutral ligand (charge $= 0$).
$H_{2}O$ is a neutral ligand (charge $= 0$).
$Cl$ is a chloride ion (charge $= -1$).
The complex is neutral,so the sum of oxidation states equals $0$.
$x + 5(0) + 1(0) + 2(-1) = 0$
$x - 2 = 0$
$x = +2$
Therefore,the oxidation number of $Ru$ is $+2$.

(D) Amorphous solids are those in which the constituent particles do not have a long-range order of arrangement.
$Glass$,$rubber$,and $plastics$ are common examples of amorphous solids.
$Camphor$ is a crystalline molecular solid.
$Magnesium$ is a metallic crystalline solid.
$Diamond$ is a covalent (network) crystalline solid.
Therefore,$Glass$ is the correct example of an amorphous solid.

(D) Amorphous solids are those in which the constituent particles do not have a long-range ordered arrangement. Examples include rubber,glass,plastic,and tar.
Crystalline solids have a long-range ordered arrangement of constituent particles. Camphor is a molecular crystalline solid.
Therefore,camphor is $NOT$ an amorphous solid.

(A) $\text{Volume of metal} = \frac{\text{Mass}}{\text{Density}} = \frac{38.6 \ g}{19.3 \ g \ cm^{-3}} = 2 \ cm^{3}$
$\text{Number of unit cells} = \frac{\text{Total volume of metal}}{\text{Volume of one unit cell}}$
$\text{Number of unit cells} = \frac{2 \ cm^{3}}{6.18 \times 10^{-23} \ cm^{3}} \approx 3.236 \times 10^{22}$

(B) The correct answer is in between $0.414$ and $0.732$.
For an ionic crystal with a coordination number of $6$,the structure is octahedral.
The radius ratio $(r_+ / r_-)$ for an octahedral geometry is mathematically derived to be in the range of $0.414$ to $0.732$.
An example of such a crystal is $NaCl$.

(A) For a body-centered cubic $(bcc)$ unit cell,the relationship between edge length $(a)$ and radius $(r)$ is given by $\sqrt{3} a = 4r$.
Given $r = 1.86 \times 10^{-8} \text{ cm}$.
$a = \frac{4r}{\sqrt{3}} = \frac{4 \times 1.86 \times 10^{-8}}{1.732} \approx 4.3 \times 10^{-8} \text{ cm}$.

(D) The radius ratio determines the type of void occupied by the cation in an ionic crystal. The limiting radius ratios for different voids are as follows:
$1$. Planar triangular void: $0.155 - 0.225$
$2$. Tetrahedral void: $0.225 - 0.414$
$3$. Octahedral void: $0.414 - 0.732$
$4$. Cubic void: $0.732 - 1.000$
Therefore,if the limiting value of $\frac{r+}{r-}$ is in the range of $0.225$ to $0.414$,it corresponds to a tetrahedral void.

(D) The packing efficiency of a $bcc$ (body-centered cubic) unit cell is $68 \%$.
The void space is calculated as $100 \% - \text{Packing Efficiency}$.
Therefore,the void space $= 100 \% - 68 \% = 32 \%$.

(C) In an $fcc$ unit cell,the number of atoms per unit cell is $Z = 4$.
The mass of the unit cell is equal to the mass of $4$ atoms.
Mass of unit cell $= 4 \times (6 \times 10^{-23} \ g) = 24 \times 10^{-23} \ g$.
Converting to scientific notation,we get $2.4 \times 10^{-22} \ g$.

(A) In an $fcc$ (face-centered cubic) unit cell,the atoms are present at the corners and the center of each face.
The relation between the edge length '$a$' and the radius of the atom '$r$' is $a = 2\sqrt{2}r$.
The number of atoms per unit cell $(Z)$ for $fcc$ is $4$.
The volume of $4$ spheres is $4 \times (\frac{4}{3} \pi r^3) = \frac{16}{3} \pi r^3$.
The volume of the unit cell is $a^3 = (2\sqrt{2}r)^3 = 16\sqrt{2}r^3$.
Packing efficiency = $\frac{\text{Volume of 4 spheres}}{\text{Volume of unit cell}} \times 100 = \frac{\frac{16}{3} \pi r^3}{16\sqrt{2}r^3} \times 100 = \frac{\pi}{3\sqrt{2}} \times 100 \approx 74.0 \%$.

(D) Volume of one unit cell $= (a)^{3} = (100 \ pm)^{3} = (100 \times 10^{-10} \ cm)^{3} = 10^{-24} \ cm^{3}$.
Total volume of $100 \ g$ of the element $= \frac{\text{Mass}}{\text{Density}} = \frac{100 \ g}{10 \ g/cm^{3}} = 10 \ cm^{3}$.
Number of unit cells $= \frac{\text{Total volume}}{\text{Volume of one unit cell}} = \frac{10 \ cm^{3}}{10^{-24} \ cm^{3}} = 1 \times 10^{25}$ unit cells.

(D) For a $bcc$ crystal,the relationship between atomic radius $r$ and edge length $a$ is given by: $r = \frac{\sqrt{3}}{4} a$
Therefore,$a = \frac{4r}{\sqrt{3}}$.
Substituting the given value $r = 1.33 \times 10^{-8} \ cm$:
$a = \frac{4 \times 1.33 \times 10^{-8} \ cm}{1.732} = \frac{5.32 \times 10^{-8} \ cm}{1.732} \approx 3.07 \times 10^{-8} \ cm$.

(D) For a $bcc$ unit cell, the number of atoms per unit cell $n = 2$.
Given density $\rho = 4 \ g \ cm^{-3}$ and edge length $a = 500 \ pm = 500 \times 10^{-10} \ cm = 5 \times 10^{-8} \ cm$.
The volume of the unit cell $a^{3} = (5 \times 10^{-8} \ cm)^{3} = 125 \times 10^{-24} \ cm^{3}$.
The formula for atomic mass $M$ is $M = \frac{\rho \times a^{3} \times N_{A}}{n}$.
Substituting the values: $M = \frac{4 \times 125 \times 10^{-24} \times 6.022 \times 10^{23}}{2} = 2 \times 125 \times 0.6022 = 150.55 \ g \ mol^{-1}$.
Thus, the atomic mass is approximately $150.5 \ g \ mol^{-1}$.

(D) For a $fcc$ unit cell,the relationship between the radius of the atom $(r)$ and the edge length of the unit cell $(a)$ is given by: $r = \frac{a}{2\sqrt{2}}$.
Given,$a = 0.3524 \ nm$.
Substituting the value of $a$ in the formula:
$r = \frac{0.3524 \ nm}{2 \times 1.414} = \frac{0.3524 \ nm}{2.828} = 0.1246 \ nm$.

(B) For an $fcc$ type unit cell, the relationship between edge length $a$ and atomic radius $r$ is given by $r = \frac{a}{2 \sqrt{2}}$.
Rearranging for $a$, we get $a = 2 \sqrt{2} \times r$.
Given $r = 127.6 \ pm$, substituting the value:
$a = 2 \times 1.414 \times 127.6 \ pm = 360.85 \ pm$.
Rounding to the nearest integer, $a \approx 361 \ pm$.

(D) The formula for density is $\rho = \frac{n \times M}{a^{3} \times N_{A}}$.
Given: $a = 4 \ \mathring{A} = 4 \times 10^{-8} \ cm$,$\rho = 2.7 \ g \ cm^{-3}$,$M = 27 \ g \ mol^{-1}$,$N_{A} = 6.022 \times 10^{23} \ mol^{-1}$.
Rearranging for $n$: $n = \frac{\rho \times a^{3} \times N_{A}}{M}$.
Substituting the values: $n = \frac{2.7 \times (4 \times 10^{-8})^{3} \times 6.022 \times 10^{23}}{27}$.
$n = \frac{2.7 \times 64 \times 10^{-24} \times 6.022 \times 10^{23}}{27} = 0.1 \times 64 \times 0.6022 \approx 3.85$.
Since the number of atoms per unit cell must be an integer,$n \approx 4$,which corresponds to a face-centered cubic $(FCC)$ unit cell.

(A) For a body-centered cubic $(BCC)$ unit cell, the relationship between the radius $(r)$ and the edge length $(a)$ is given by $r = \frac{\sqrt{3} a}{4}$.
Given that the edge length $a = 351 \ pm$.
Substituting the value of $a$ into the formula:
$r = \frac{\sqrt{3} \times 351}{4} = \frac{1.732 \times 351}{4} = \frac{607.932}{4} = 151.98 \ pm$.

(B) For a face-centred cubic $(fcc)$ unit cell, the relationship between the edge length $(a)$ and the atomic radius $(r)$ is given by $r = \frac{a}{2 \sqrt{2}}$.
Therefore, the edge length $a = 2 \sqrt{2} \times r$.
Given $r = 125 \text{ pm}$, we have $a = 2 \times 1.414 \times 125 \text{ pm} = 353.5 \text{ pm}$.

(D) Given: Edge length $a = 250 \ pm = 2.5 \times 10^{-8} \ cm$.
Atomic mass $M = 90.3 \ g \ mol^{-1}$.
For $fcc$ lattice,the number of atoms per unit cell $n = 4$.
Avogadro's number $N_{A} = 6.022 \times 10^{23} \ mol^{-1}$.
The density formula is $\rho = \frac{n \times M}{a^{3} \times N_{A}}$.
Substituting the values: $\rho = \frac{4 \times 90.3}{(2.5 \times 10^{-8})^{3} \times 6.022 \times 10^{23}}$.
$\rho = \frac{361.2}{15.625 \times 10^{-24} \times 6.022 \times 10^{23}} = \frac{361.2}{9.409} \approx 38.40 \ g \ cm^{-3}$.

(C) Given: Density $(\rho)$ = $2.8 \ g \ cm^{-3}$,Edge length $(a)$ = $4 \times 10^{-8} \ cm$,Number of atoms per unit cell $(n)$ = $4$ (for $fcc$),Avogadro's number $(N_{A})$ = $6.022 \times 10^{23} \ mol^{-1}$.
Formula: $\rho = \frac{n \times M}{a^{3} \times N_{A}}$.
Rearranging for molar mass $(M)$: $M = \frac{\rho \times a^{3} \times N_{A}}{n}$.
Substituting the values: $M = \frac{2.8 \times (4 \times 10^{-8})^{3} \times 6.022 \times 10^{23}}{4}$.
$M = \frac{2.8 \times 64 \times 10^{-24} \times 6.022 \times 10^{23}}{4}$.
$M = 0.7 \times 64 \times 0.6022 = 26.97896 \approx 27 \ g \ mol^{-1}$.

(A) Edge length $a = 16 \times 10^{-8} \ cm$.
Volume of one unit cell $= a^{3} = (16 \times 10^{-8} \ cm)^{3} = 4096 \times 10^{-24} \ cm^{3}$.
Since the unit cell contains $24$ molecules,the volume occupied by $24$ molecules is $4096 \times 10^{-24} \ cm^{3}$.
Volume of $1$ mole of molecules $= (\text{Volume of } 24 \text{ molecules} / 24) \times N_{A}$.
Volume of $1$ mole $= (4096 \times 10^{-24} / 24) \times 6.022 \times 10^{23} \ cm^{3} \ mol^{-1}$.
Volume of $1$ mole $\approx 170.67 \times 10^{-24} \times 6.022 \times 10^{23} \approx 102.77 \ cm^{3} \ mol^{-1}$.

(A) For a $bcc$ unit cell,the relationship between the radius $r$ and the edge length $a$ is given by $r = \frac{\sqrt{3} a}{4}$.
Rearranging the formula to solve for $a$,we get $a = \frac{4 r}{\sqrt{3}}$.
Substituting the given value $r = 1.86 \times 10^{-8} \ cm$:
$a = \frac{4 \times 1.86 \times 10^{-8} \ cm}{1.732} = 4.29 \times 10^{-8} \ cm$.

(B) Given edge length $a = 300 \ pm = 300 \times 10^{-10} \ cm = 3 \times 10^{-8} \ cm$.
For a $bcc$ unit cell, the relationship between radius $r$ and edge length $a$ is given by $r = \frac{\sqrt{3} a}{4}$.
Substituting the values: $r = \frac{1.732 \times 3 \times 10^{-8} \ cm}{4}$.
$r = \frac{5.196 \times 10^{-8} \ cm}{4} = 1.299 \times 10^{-8} \ cm$.

(C) In a cubic close packing $(ccp)$ or face-centered cubic $(fcc)$ structure,each atom is surrounded by $12$ nearest neighbors.
Specifically,an atom at a face center is in contact with $4$ atoms in its own plane,$4$ atoms in the plane above,and $4$ atoms in the plane below.
Therefore,the coordination number is $12$.

(C) In an ionic crystal lattice,the coordination number of an ion is determined by the type of void (hole) it occupies.
For a cubic void,the cation is surrounded by $8$ anions located at the corners of a cube.
Therefore,the coordination number of the cation in a cubic hole is $8$.

(B) $1$. Edge length $a = 3 \mathring{A} = 3 \times 10^{-8} \ cm$.
$2$. Volume of the unit cell $V = a^{3} = (3 \times 10^{-8} \ cm)^{3} = 27 \times 10^{-24} \ cm^{3}$.
$3$. Mass of one unit cell = $\text{Volume} \times \text{Density} = 27 \times 10^{-24} \ cm^{3} \times 8 \ g/cm^{3} = 216 \times 10^{-24} \ g$.
$4$. Number of unit cells in $100 \ g = \frac{100 \ g}{216 \times 10^{-24} \ g/\text{unit cell}} = 0.4629 \times 10^{24} = 4.63 \times 10^{23}$ unit cells.

(A) For an $fcc$ (face-centered cubic) unit cell, the relationship between the edge length $(a)$ and the atomic radius $(r)$ is given by: $a = 2 \sqrt{2} r$ or $r = \frac{a}{2 \sqrt{2}}$.
Given edge length $a = 620 \ pm$.
Substituting the value: $r = \frac{620}{2 \times 1.414} = \frac{620}{2.828} \approx 219.2 \ pm$.
Thus, the radius of the $Xe$ atom is $219.2 \ pm$.

(B) For $fcc$ structure,the number of atoms per unit cell $Z = 4$.
Density $d = \frac{Z \cdot M}{a^3 \cdot N_A}$.
Rearranging for edge length $a$: $a^3 = \frac{Z \cdot M}{d \cdot N_A}$.
Substituting the values: $a^3 = \frac{4 \times 98.99}{3.4 \times 6.022 \times 10^{23}}$.
$a^3 = 193.389 \times 10^{-24} \text{ cm}^3$.
$a = \sqrt[3]{193.389} \times 10^{-8} \text{ cm} = 5.783 \times 10^{-8} \text{ cm}$.
Since $1 \text{ cm} = 10^8 \mathring A$,$a = 5.783 \mathring A$.

$(A)$ For a $bcc$ unit cell, the number of atoms per unit cell $(n)$ is $2$.
The edge length $(a)$ is $288 \ pm = 288 \times 10^{-10} \ cm = 2.88 \times 10^{-8} \ cm$.
The density formula is $\rho = \frac{n \times M}{a^{3} \times N_{A}}$.
Rearranging for molar mass $(M)$: $M = \frac{\rho \times a^{3} \times N_{A}}{n}$.
Substituting the values: $M = \frac{7.2 \times (2.88 \times 10^{-8})^{3} \times 6.022 \times 10^{23}}{2}$.
$M = \frac{7.2 \times 23.8878 \times 10^{-24} \times 6.022 \times 10^{23}}{2} \approx \frac{103.56}{2} \approx 51.78 \ g \ mol^{-1}$.
Thus, the correct option is $A$.

(B) For a face-centered cubic $(fcc)$ unit cell, the relationship between the edge length $(a)$ and the atomic radius $(r)$ is given by: $a = 2 \sqrt{2} r$ or $r = \frac{a}{2 \sqrt{2}}$.
Given edge length $a = 316.5 \text{ pm}$.
Substituting the value: $r = \frac{316.5 \text{ pm}}{2 \times 1.414} = \frac{316.5 \text{ pm}}{2.828} \approx 111.91 \text{ pm}$.
Therefore, the correct option is $B$.

(B) For a $bcc$ lattice, the number of atoms per unit cell, $n = 2$.
Given: edge length $a = 500 \ pm = 5 \times 10^{-8} \ cm$, density $\rho = 7.5 \ g \ cm^{-3}$, mass $m = 300 \ g$.
Volume of unit cell $a^{3} = (5 \times 10^{-8} \ cm)^{3} = 125 \times 10^{-24} \ cm^{3}$.
Using the density formula $\rho = \frac{nM}{a^{3} N_{A}}$, we find the molar mass $M$:
$M = \frac{\rho \times a^{3} \times N_{A}}{n} = \frac{7.5 \times 125 \times 10^{-24} \times 6.022 \times 10^{23}}{2} \approx 282.3 \ g \ mol^{-1}$.
Number of atoms in $300 \ g$ of metal = $\frac{\text{mass}}{\text{molar mass}} \times N_{A} = \frac{300}{282.3} \times 6.022 \times 10^{23} \approx 6.4 \times 10^{23} \ \text{atoms}$.

(B) In a $bcc$ (body-centered cubic) unit cell,atoms are present at all $8$ corners and one at the center of the body.
Each corner atom is shared by $8$ adjacent unit cells.
Therefore,the contribution of each atom at the corner to a single unit cell is $1/8$.

(B) The coordination number of an octahedral void is $6$.
Therefore,if a cation occupies an octahedral hole,its coordination number is $6$.

(C) The correct answer is $CsCl$.
In the $CsCl$ crystal lattice,the $Cl^{-}$ ions form a simple cubic arrangement,and the $Cs^{+}$ cation occupies the body-centered cubic hole.
Since there is only one body-centered hole per unit cell,the $Cs^{+}$ ion occupies this single cubic hole.
Therefore,$CsCl$ is the crystal lattice where the cation occupies the cubic hole.

(A) The volume of the given sample is calculated as: $\text{Volume} = \frac{\text{Mass}}{\text{Density}} = \frac{36 \ g}{7.2 \ g \ cm^{-3}} = 5 \ cm^{3}$.
The number of unit cells in the sample is given by the ratio of the total volume of the sample to the volume of one unit cell:
$\text{Number of unit cells} = \frac{\text{Total Volume}}{\text{Volume of one unit cell}} = \frac{5 \ cm^{3}}{24.99 \times 10^{-24} \ cm^{3}}$.
$\text{Number of unit cells} \approx 0.20008 \times 10^{24} = 2.0 \times 10^{23}$.

(B) For an $fcc$ unit cell,the packing efficiency is $74 \%$.
Therefore,the percentage of unoccupied space (void space) is $100 \% - 74 \% = 26 \%$.

(B) The number of atoms per $bcc$ type unit cell is $n = 2$.
Mass of $bcc$ unit cell $=$ Mass of $2$ atoms.
$\therefore$ Mass of $bcc$ unit cell of $Na = 3.819 \times 10^{-23} \ g \times 2 = 7.638 \times 10^{-23} \ g$.

(B) The radius ratio is defined as the ratio of the radius of the cation to the radius of the anion: $\text{Radius ratio} = \frac{r_+}{r_-}$.
Given,$\text{Radius ratio} = 0.5248$ and $r_+ = 0.95 \ \mathring{A}$.
Substituting the values: $0.5248 = \frac{0.95}{r_-}$.
Therefore,$r_- = \frac{0.95}{0.5248} \approx 1.81 \ \mathring{A}$.

(D) Given that the radius of the anion $(r_a)$ is double the radius of the cation $(r_c)$,so $r_a = 2r_c$.
The radius ratio is calculated as $\frac{r_c}{r_a} = \frac{r_c}{2r_c} = 0.5$.
For a radius ratio in the range $0.414 - 0.732$,the coordination number is $6$ and the cation occupies an octahedral hole.

(C) $O_2$ is paramagnetic.
Benzene and water are diamagnetic.
$Fe$ (Iron) is ferromagnetic.

(B) $Gadolinium$ is ferromagnetic in nature.
$Oxygen$ is paramagnetic.
$Water$ and $Benzene$ are diamagnetic.

(D) true solution is a homogeneous mixture of two or more substances.
It is not a heterogeneous mixture.
In a true solution,the particle size of the solute is less than $10^{-9} \ m$ or $1 \ nm$.
Since the components are uniformly distributed at the molecular level,it exhibits consistent properties throughout.
Therefore,the statement in option $D$ is incorrect.

(B) Molar mass of $NaOH = 23 + 16 + 1 = 40 \ g \ mol^{-1}$.
Number of moles of $NaOH = \frac{\text{mass}}{\text{molar mass}} = \frac{2.0 \ g}{40 \ g \ mol^{-1}} = 0.05 \ mol$.
Volume of solution $= 500 \ cm^{3} = 0.5 \ L$.
Molarity $(M) = \frac{\text{moles of solute}}{\text{volume of solution in } L} = \frac{0.05 \ mol}{0.5 \ L} = 0.1 \ mol \ dm^{-3}$.

(A) Given: Mass of solute $(CH_3COOH)$ $= 60 \ g$,Volume of solution $= 1 \ dm^3 = 1000 \ cm^3$,Density of solution $= 1.25 \ g / cm^3$.
Molar mass of $CH_3COOH = 12 + 3 + 12 + 16 + 16 + 1 = 60 \ g / mol$.
Moles of solute $= \frac{60 \ g}{60 \ g / mol} = 1 \ mol$.
Mass of solution $= \text{Volume} \times \text{Density} = 1000 \ cm^3 \times 1.25 \ g / cm^3 = 1250 \ g$.
Mass of solvent $= \text{Mass of solution} - \text{Mass of solute} = 1250 \ g - 60 \ g = 1190 \ g = 1.19 \ kg$.
Molality $(m) = \frac{\text{Moles of solute}}{\text{Mass of solvent in kg}} = \frac{1 \ mol}{1.19 \ kg} \approx 0.84 \ m$.
Note: Assuming the $1 \ dm^3$ refers to the volume of the solution,the calculated molality is approximately $0.84 \ m$. Given the options,$0.8 \ m$ is the closest approximation.

(C) The vapour pressure of a solution is determined by the mole fraction of the solvent. According to Raoult's Law,$P_{sol} = X_{solvent} \times P^0_{solvent}$.
Adding water to an aqueous solution increases the amount of solvent relative to the solute,thereby increasing the mole fraction of the solvent $(X_{solvent})$.
Since $P^0_{solvent}$ (vapour pressure of pure water) is constant at a given temperature,an increase in $X_{solvent}$ leads to an increase in the vapour pressure of the solution.
Adding more solute (as in options $A$,$B$,and $D$) would decrease the mole fraction of the solvent,thereby decreasing the vapour pressure.

(C) Step $1$: Calculate the number of moles of solute $(n_2)$ and solvent $(n_1)$.
$n_2 = \frac{38.4 \ g}{384 \ g \ mol^{-1}} = 0.1 \ mol$
$n_1 = \frac{116 \ g}{58 \ g \ mol^{-1}} = 2 \ mol$
Step $2$: Calculate the mole fraction of the solvent $(x_1)$.
$x_1 = \frac{n_1}{n_1 + n_2} = \frac{2}{2 + 0.1} = \frac{2}{2.1} \approx 0.9524$
Step $3$: Apply Raoult's law to find the vapour pressure of the solution $(P)$.
$P = x_1 \times P_0 = 0.9524 \times 0.842 \ atm \approx 0.8019 \ atm$.
Note: Based on the provided options,the closest value is $0.7999 \ atm$.

(A) According to Henry's law,the solubility $(S)$ of a gas is given by the formula: $S = K_{H} \times P$
Given: $K_{H} = 1.3 \times 10^{-3} \ mol \ dm^{-3} \ atm^{-1}$ and $P = 0.46 \ atm$
Substituting the values: $S = (1.3 \times 10^{-3} \ mol \ dm^{-3} \ atm^{-1}) \times (0.46 \ atm) = 5.98 \times 10^{-4} \ mol \ dm^{-3}$