The elevation of boiling point of 0.10 m aqueous CrCl_3.xNH_3 solution is two times that of 0.05 m aqueous CaCl_2 solution. The value of x is \underline\hspace2cm . [Assume 100% ionisation of the complex and CaCl_2 , coordination number of Cr as 6, and that all NH_3 molecules are present inside the coordination sphere]

Solutions PYQ — JEE Main 2020 · 6 Sep · Shift 1 · Q2 with Solution

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Correct answer: 5

For dilute solutions, $\Delta T_b = i K_b m$ where:

Since both solutions are aqueous, $K_b$ is same.

Given: $\Delta T_b(\text{complex solution}) = 2\,\Delta T_b(\text{CaCl}_2\text{ solution})$

So, $i_1 m_1 = 2(i_2 m_2)$

Assuming 100% ionisation: $CaCl_2 \rightarrow Ca^{2+} + 2Cl^-$ Total particles $=3$

Hence, $i_2 = 3$

Given molality: $m_2 = 0.05$

Thus, $i_2 m_2 = 3 \times 0.05 = 0.15$

Therefore, $i_1 m_1 = 2 \times 0.15 = 0.30$

Given for complex solution: $m_1 = 0.10$

So, $i_1 \times 0.10 = 0.30$ $i_1 = 3$

Given compound: $CrCl_3\cdot xNH_3$

All $NH_3$ are inside coordination sphere and coordination number of Cr is 6.

So the complex must be of the form: $[Cr(NH_3)_xCl_{6-x}]Cl_{3-(6-x)}$

Because total 3 chlorides are present, some are inside the coordination sphere and the rest are outside as counter ions.

More simply, if $x$ ammonia molecules are coordinated, then number of coordinated chlorides is: $6-x$

Hence outside chlorides: $3-(6-x)=x-3$

Thus the compound ionises as: $[Cr(NH_3)_xCl_{6-x}]Cl_{x-3} \rightarrow [Cr(NH_3)_xCl_{6-x}]^{(x-3)+} + (x-3)Cl^-$

Total number of ions produced: $i_1 = 1 + (x-3) = x-2$

So, $x-2 = 3$ $x = 5$

$\boxed{5}$

Stored correct answer = 5, which matches the derived answer.