Question

0.012 g of a divalent metal is completely dissolved in 40 cc of N/10 HCL. The exces of acid required 15cc of N/5 NaOH for neutralization Find
the atomic weight of the metal.​

Answer 1

Answer:

same doubt

Explanation:

if you come to know please tell me also

Answer 2

Answer:24 amu

Explanation:

For HCl, volume (Va)=40cc

Normality (Na)= N/10=0.1

For metal, weight (w)=0.012g

Valency (n)=2 (divalent is given in question)

For NaOH, Volume (Vb)=15cc

Normality (NB)=N/5=0.2

Now, for complete neutralization

No. of gm equivalent of HCl=no.of gm equivalent of (metal + NaOH)

Or, (Va*Na)/1000= w/equivalent wt. +(Vb*NB)/1000

Or,(40*0.1)*1000= 0.012/eq.wt +(15*0.2)*1000

Or,0.001 eq.wt=0.012

Or, Eq.wt=12

Here,

Eq .wt= Atomic wt/ valency (n)

Therefore, atomic wt= 12*2=24 amu

Hope ❤️ this will be helpful for you ...Thanks