SATHEE: Solutions Question 360

Solutions Question 360

Question: 0.010M solution an acid HA freezes at $- 0.0205^{\circ} C$ . If $K_{f}$ for water is $1.860 K k g m o l^{- 1}$ , the ionization constant of the conjugate base of the acid will be (assume $0.010 M = 0.010 m$ )

Options:

A) $1.1 \times 10^{- 4}$

B) $1.1 \times 10^{- 3}$

C) $9.0 \times 10^{- 11}$

D) $9.0 \times 10^{- 12}$

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Answer:

Correct Answer: C

Solution:

$Δ T_{f} (n o r m a l) = K_{f} m = 1.86 \times 0.01 = 0.0186;$

$i = \frac{Δ T_{f (o b s)}}{Δ T_{f (n o r)}} = \frac{0.0205}{0.0186} = 1.10 = 1 + α;$

$α = 0.1$

$K_{a} = \frac{C a^{2}}{1 - α} = \frac{0.01 \times {0.1}^{2}}{1 - 0.1} = \frac{1}{9} \times 10^{- 3};$

$K_{a} = \frac{K_{w}}{K_{a}} = 10 \times 10^{- 14} \times 9 \times 10^{3} = 9 \times 10^{- 11}$

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Solutions Question 360

Question: 0.010M solution an acid HA freezes at $- 0.0205^{\circ} C$ . If $K_{f}$ for water is $1.860 K k g m o l^{- 1}$ , the ionization constant of the conjugate base of the acid will be (assume $0.010 M = 0.010 m$ )

Options:

Answer:

Solution:

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Solutions Question 360

Question: 0.010M solution an acid HA freezes at −0.0205∘C . If Kf for water is 1.860Kkgmol−1 , the ionization constant of the conjugate base of the acid will be (assume 0.010M=0.010m )

Options:

Answer:

Solution:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

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Question: 0.010M solution an acid HA freezes at $- 0.0205^{\circ} C$ . If $K_{f}$ for water is $1.860 K k g m o l^{- 1}$ , the ionization constant of the conjugate base of the acid will be (assume $0.010 M = 0.010 m$ )