Select the true statements about hydrocarbons.
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A 25.00 ml sample of hydrochloric acid solution, HCl, is titrated with 0.0512 m NaOH solution. the volume of NaOH solution required is 21.68 ml then the molarity of the HCl solution is 0.044 M .
Calculation ,
Formula used : ...( i )
Where M is the molarity or concentration and V is the volume in ml .
concentration of hydrochloric acid solution ( ) = ?
concentration of NaOH ( ) = 0.0512 M
volume of hydrochloric acid solution ( ) = 25.00 ml
volume of NaOH ( ) = 21.68 ml
Putting the value of concentration , volume of both in equation ( i ) we get .
× 25.00 ml = 0.0512 × 21.68 ml
= 0.0512 × 21.68 ml / 25.00 ml= 0.044 M
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