किसी (0.1,m) विलयन का हिमांक अवनमन (0.279,K) है। यदि \(K_f=1.86,K,kg,mol^{-1}\), तो (i) क्या होगा?
A (0.1,m) solution has freezing point depression (0.279,K). If \(K_f=1.86,K,kg,mol^{-1}\), what is (i)?
Explanation opens after your attempt
C. (1.5)
Concept
\(\Delta T_f=iK_fm\) से \(i=\frac{\Delta T_f}{K_fm}\)। / From \(\Delta T_f=iK_fm\), \(i=\frac{\Delta T_f}{K_fm}\).
Why this answer is correct
\(i=\frac{0.279}{1.86\times0.1}=\frac{0.279}{0.186}=1.5\)। / \(i=\frac{0.279}{1.86\times0.1}=\frac{0.279}{0.186}=1.5\).
Exam Tip
इकाई सही रखने से गणना छोटी और स्पष्ट होती है। / Keeping units correct makes the calculation short and clear.
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