Complex A has a composition of $$H_{12}O_6Cl_3Cr$$. If the complex on treatment with conc. $$H_2SO_4$$ loses 13.5% of its original mass, the correct molecular formula of A is:
[Given: atomic mass of Cr = 52 amu and Cl = 35 amu]

We are told that the complex A has the elemental composition $$H_{12}O_6Cl_3Cr$$ and that, on treatment with concentrated $$H_2SO_4$$, it loses exactly $$13.5\%$$ of its original mass. Concentrated sulphuric acid is a strong dehydrating agent, so it removes only those water molecules that are present outside the coordination sphere (called “water of crystallisation”). Any water molecule directly bonded to the metal as a ligand remains attached and therefore is not lost. Hence, the observed percentage loss of mass must correspond precisely to the mass of lattice water present in one mole of the complex.

We now test each of the four candidate formulae one by one, computing (a) the molar mass of the entire complex, (b) the mass of the lattice water molecules present and (c) the percentage loss obtained by dividing these two quantities. Throughout the calculation we use the given atomic masses $$\bigl(M_{Cr}=52\,\text{amu},\;M_{Cl}=35\,\text{amu},\;M_H=1\,\text{amu},\;M_O=16\,\text{amu}\bigr)$$ and the well-known molar mass of water $$M_{H_2O}=2(1)+16=18\,\text{amu}.$$ We also remember that the molar mass of the complex is simply the sum of the individual atomic masses appearing in its formula.

Option A: $$[Cr(H_2O)_6]Cl_3$$

Step 1 - Molar mass

$$M\bigl([Cr(H_2O)_6]Cl_3\bigr)=52+6(18)+3(35)=52+108+105=265\ \text{amu}.$$

Step 2 - Lattice water present: none (all six waters are ligands).

Step 3 - Percentage loss

$$\%\,\text{loss}=\frac{0}{265}\times100=0\%.$$

This does not match the required $$13.5\%$$, so Option A is rejected.

Option B: $$[Cr(H_2O)_3Cl_3]\cdot3H_2O$$

Step 1

$$M\bigl([Cr(H_2O)_3Cl_3]\cdot3H_2O\bigr)=52+3(18)+3(35)+3(18)=52+54+105+54=265\ \text{amu}.$$

Step 2

$$\text{Lattice water mass}=3(18)=54\ \text{amu}.$$

Step 3

$$\%\,\text{loss}=\frac{54}{265}\times100\approx20.4\%.$$

This is higher than $$13.5\%,$$ so Option B is rejected.

Option C: $$[Cr(H_2O)_5Cl]Cl_2\cdot H_2O$$

Step 1

$$M\bigl([Cr(H_2O)_5Cl]Cl_2\cdot H_2O\bigr)=52+5(18)+1(35)+2(35)+1(18)=52+90+35+70+18=265\ \text{amu}.$$

Step 2

$$\text{Lattice water mass}=1(18)=18\ \text{amu}.$$

Step 3

$$\%\,\text{loss}=\frac{18}{265}\times100\approx6.8\%.$$

This is lower than $$13.5\%,$$ so Option C is also rejected.

Option D: $$[Cr(H_2O)_4Cl_2]Cl\cdot2H_2O$$

Step 1

$$M\bigl([Cr(H_2O)_4Cl_2]Cl\cdot2H_2O\bigr)=52+4(18)+2(35)+1(35)+2(18)=52+72+70+35+36=265\ \text{amu}.$$

Step 2

$$\text{Lattice water mass}=2(18)=36\ \text{amu}.$$

Step 3

$$\%\,\text{loss}=\frac{36}{265}\times100\approx13.6\%.$$

The calculated percentage loss $$\bigl(13.6\%\bigr)$$ is in excellent agreement with the observed value of $$13.5\%.$$ Therefore, Option D is the only formula that correctly reproduces the experimental mass loss.

Hence, the correct answer is Option 4.