JEE Advanced · Chemistry · 16. Solutions
The boiling point of water in a \(0.1\) molal silver nitrate solution (solution \(\mathbf{A}\) ) is \(\mathbf{x}^{\circ} \mathrm{C}\). To this solution \(\mathbf{A}\), an equal volume of \(0.1\) molal aqueous barium chloride solution is added to make a new solution B. The difference in the boiling points of water in the two solutions \(\mathbf{A}\) and \(\mathbf{B}\) is \(\mathbf{y} \times 10^{-2}{ }^{\circ} \mathrm{C}\).
(Assume: Densities of the solutions \(\mathbf{A}\) and \(\mathbf{B}\) are the same as that of water and the soluble salts dissociate completely.
Use: Molal elevation constant (Ebullioscopic Constant), \(K_{b}=0.5 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\); Boiling point of pure water as \(100^{\circ} \mathrm{C}\).)
The value of is ___.
- A 111.1
- B 100.1
- C 122
- D 152
Answer & Solution
Correct Answer
(B) 100.1
Step-by-step Solution
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