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Question: The minimum volume of water required to dissolve 0.1g lead(II) chloride to get a saturated solution....

The minimum volume of water required to dissolve 0.1g lead(II) chloride to get a saturated solution. Ksp{K_{sp}}of PbCl2=3.2×10−8PbC{l_2} = 3.2 \times {10^{ - 8}};atomic mass ofPb=207uPb = 207u
A.0.18L0.18L
B.17.98L17.98L
C.1.798L1.798L
D.0.36L0.36L

Explanation

Solution

Solubility of a substance is the maximum amount that can be dissolved in a particular amount of solvent. The solubility product is the constant level at which a solute dissolves in the solution. It depends on the solute, solvent, temperature and pressure.

Formula used:

K=[C]c[D]d[A]a Ksp=[C]c[D]d  K = \dfrac{{{{\left[ C \right]}^c}{{\left[ D \right]}^d}}}{{{{\left[ A \right]}^a}}} \\\ {K_{sp}} = {\left[ C \right]^c}{\left[ D \right]^d} \\\

Complete step by step solution: We know that a solution is a homogeneous mixture of two or more components which states that the different components have the same composition. The composition of a solution can be estimated by calculating the concentration of the solution and the concentration of a solution can be expressed either qualitatively or quantitatively.
Solute+Solvent⇌SolutionSolute + Solvent \rightleftharpoons Solution
It is given to us that,
Ksp{K_{sp}}of PbCl2=3.2×10−8PbC{l_2} = 3.2 \times {10^{ - 8}}
Considering the equation,
PbCl2⇌Pb2++2Cl−PbC{l_2} \rightleftharpoons P{b^{2 + }} + 2C{l^ - }
According to the formula of solubility product,
Ksp=(S)(2S)2{K_{sp}} = \left( S \right){\left( {2S} \right)^2}that means
⇒Ksp=4S3\Rightarrow {K_{sp}} = 4{S^3}
Solving this to get the value of S,

4S3=3.2×10−8 ⇒S3=3.2×10−84 ⇒S3=8×10−9 ⇒S=2×10−3  4{S^3} = 3.2 \times {10^{ - 8}} \\\ \Rightarrow {S^3} = \dfrac{{3.2 \times {{10}^{ - 8}}}}{4} \\\ \Rightarrow {S^3} = 8 \times {10^{ - 9}} \\\ \Rightarrow S = 2 \times {10^{ - 3}} \\\

Molecular mass ofPbCl2=278gmol−1PbC{l_2} = 278gmo{l^{ - 1}}
Now on using Molarity Formula,
S=weightMolecularMass×volumeS = \dfrac{{weight}}{{MolecularMass \times volume}}
WM×V=2×10−3\dfrac{W}{{M \times V}} = 2 \times {10^{ - 3}}
Substituting the values
0.1278×V=2×10−3\dfrac{{0.1}}{{278 \times V}} = 2 \times {10^{ - 3}}
Solving it to get the value of V
V=100278×2V = \dfrac{{100}}{{278 \times 2}}
We get
V=0.18LV = 0.18L

So, the correct answer to the given question is option A(0.18L0.18L).

Note: The polar solutes dissolve in polar solvents and the non-polar solutes dissolve in non-polar solvents. They dissolve if the intermolecular spaces are similar thus, like dissolves like. For an endothermic reaction, solubility increases and for an exothermic reaction solubility decreases. The solubility of gases increases with increase in pressure.