SATHEE: Solutions Question 56

Solutions Question 56

Question: A 0.001 molal solution of $ [Pt{{(NH_3)}4}Cl_4] $i n w a t e r h a d a f r e e z i n g p o i n t d e p r e s s i o n o f$ {{0.0054}^{o}}C $. I f$ K{f} $ for water is 1.80, the correct formulation for the above molecule is [Kerala CET (Med.) 2003]

Options:

A) $[P t {(N H_{3})}_{4} C l_{3}], C l$

B) $[P t {(N H_{3})}_{4} C l], C l_{2}$

C) $[P t {(N H_{3})}_{4} C l_{2}], C l_{3}$

D) $[P t {(N H_{3})}_{4} C l_{4}]$

Show Answer

Answer:

Correct Answer: B

Solution:

$Δ T_{f} = i m k_{f}$ ; $0.0054 = i \times 1.8 \times 0.001$

$i = 3$ so it is $[P t {(N H_{3})}_{4} C l] C l_{2}$ .

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Solutions Question 56

Question: A 0.001 molal solution of $ [Pt{{(NH_3)}4}Cl_4] inwaterhadafreezingpointdepressionof {{0.0054}^{o}}C .If K{f} $ for water is 1.80, the correct formulation for the above molecule is [Kerala CET (Med.) 2003]

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Question: A 0.001 molal solution of $ [Pt{{(NH_3)}4}Cl_4] $i n w a t e r h a d a f r e e z i n g p o i n t d e p r e s s i o n o f$ {{0.0054}^{o}}C $. I f$ K{f} $ for water is 1.80, the correct formulation for the above molecule is [Kerala CET (Med.) 2003]