250 mL  of 0.5 M NaOH was added to 500 mL of 1 M HCl. The number of unreacted HCl molecules in the solution after complete reaction is_________ * 1021 (Nearest integer)

(NA = 6.022 *1023)

4 Views|Posted 7 months ago
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7 months ago

NaOH      +      HCl   ->  NaCl     +      H2O

(milimole) t = 0                250 * 0.5     500 * 1.0     0            0           

 t =        -         375        125        125 millimole

unreacted HCl mole = 3 7 5 1 0 0 0  

number of HCl molecule unreacted = 3 7 5 1 0 0 0 * 6 . 0 2 3 * 1 0 2 3  

= 225.75 * 1021 226 * 1021

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Chemistry Ncert Solutions Class 11th 2023

Chemistry Ncert Solutions Class 11th 2023

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