JEE Mains · Chemistry · STD 12 -1. Solution and colligative properties
Freezing point of a \(4\%\) aqueous solution of \(X\) is equal to freezing point of \(12\%\) aqueous solution of \(Y\). If molecular weight of \(X\) is \(A\) then molecular weight of \(Y\) is ............. \(\mathrm{A}\)
- A \(3\)
- B \(2\)
- C \(1\)
- D \(4\)
Answer & Solution
Correct Answer
(A) \(3\)
Step-by-step Solution
Detailed explanation
\({(\Delta {T_f})_Y}\, = \,{(\Delta {T_f})_Y}\) \({k_f}{m_X}\, = \,{k_t}{m_Y}\) \(\frac{{4\, \times \,1000}}{{A\, \times \,96}}\, = \,\frac{{12\, \times \,1000}}{{M\, \times \,88}}\) \(M\, = \,3.27\,A\, = \,3\,A\)
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