JEE Mains · Chemistry · STD 12 - 8.2 Carboxylic acids and their derivative
\(150\,g\) of acetic acid was contaminated with \(10.2\,g\) ascorbic acid \(\left( C _{6} H _{8} O _{6}\right)\) to lower down its freezing point by \(\left( x \times 10^{-1}\right)^{\circ} C\). The value of \(x\) is (Nearest integer) [Given \(K _{ f }=3.9\,K\,kg\,mol ^{-1}\);Molar mass of ascorbic acid \(=176\,g\,mol ^{-1}\) ]
- A \(14\)
- B \(13\)
- C \(15\)
- D \(11\)
Answer & Solution
Correct Answer
(C) \(15\)
Step-by-step Solution
Detailed explanation
\(150\,g CH _{3} COOH\) \(10.2\,g\) ascorbic acid \(\Rightarrow 0.058\) moles \(\Delta T _{ f }=\left( x \times 10^{-1}\right)^{\circ} C\) \(\Delta T _{ f }= k _{ f } \cdot \text { molality }\) \(=3.9 \times \frac{0.058}{150} \times 1000\) \(=1.5^{\circ}\,C\)…
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