Question

0.5g of Ba(OH)2,0.01molbarium hydroxide and 0.01aq barium hydroxide were diluted together to one litre. calculated the normality of basic solution​

Answer 1

Weight Ba(OH)2=0.5 g

Molwt Ba(OH)2=Ba +2(O)+2(H)

= 137+32+2=171

Moles=weight /molecular weight

Moles = 0.5/171=0.003

Volume= 1 litre

N Factor for Ba(OH)2 is 2

Normality= Molarity ×n Factor

Molarity= moles of[Ba(OH)2]/volume

Molarity=0.003/1=0.003

Normality= 0.003×2=0.006N

I hope this helps (θ‿θ)

Answer 2

0.0058N =0.006N (approx)

Explanation:

wt of Ba(OH) 2=0.5g

Molar mass of Ba(OH) 2=137(Ba),16(O),1(H)

=137+32+2=171g

Gram equivalent weight=molarmass/no.of replacable OH-ions

171/2 =85.5g

N=wt/Gew×1/V

N=0.5/85.5×1/1

N=0.058 =0.006(approx)