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If 145 mL of 1.00 M KOH is diluted to 1.15 L, the resulting solution contains O 2.30 moles of KOH O 0.29 moles of KOH O 0.145
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0 given data :- initial veume of KOH= 145 m2 (ر) -0.145L (11 =1000mL) initial concentrahcm, ot ROH = 1M final roume (,) =1010.05 0.209 X 80 mL XM Too = oros X80 mL = 40mL To E umL so um noume required to prepare 0.20m Naoh. so option is correct 0 (B​​​​​​​

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