100ml solution of an acid (M.W. = 82) containing 39g of acid per litre was completely neutrlaized by 95mL of
aqueous NaOH solution containing 40g of NaOH per litre. The basicity of acid is
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Answer:no. Of moles of NAOH= 40g/40gmol- = 1mol
No. Of moles of acid= 39/82
For neutralisation,
NI VI =N2 V2
(Molarity)1 × acidity × VI = (Molarity)2 ×basicity × V2
=No.of moles of NAOH/1l X100= no.of moles of acid/1l × acidity× V2
=1/1×100= 32/82×basicity×95
Basicity= (82×100)/(95×39)
=2.21
=2approximately
Explanation:
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