At 90- C pure water has - H 3 O -10-6 mol litre -1- The value of K W at 90- C isA- 10-6B- 10-12C- 10-14D- 10-8

The correct option is B Say x = [H_3 O^+]=[OH^ 1] =10^ 6mol litre^ 1 K_W= [H_3 O^+] × [OH^ 1] 10^ 12 (mol litre^ 1)^ 2 ...

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At 90∘ C pure...

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At $90^{\circ}$ C pure water has [ ${H_{3} O}^{+}$ ] = $10^{- 6}$ mol ${l i t r e}^{- 1}$ . The value of $K_{W}$ at $90^{\circ}$ C is

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At $90^{\circ}$ C pure water has [ ${H_{3} O}^{+}$ ] = $10^{- 6}$ mol ${l i t r e}^{- 1}$ . The value of $K_{W}$ at $90^{\circ}$ C is

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[H_{3} O^{+}] = 10^{- 6} m o l L^{- 1} .

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[H_{3} O^{+}] = 10^{- 6.5}

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At 363K, pure water has $[H_{3} O^{+}] = 10^{- 6} M$ . The value of Kw at this temperature will be

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[H^{+}] = 10^{- 6} M

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