Question

A metal weighing 0.43 gram was dissolved in 50 ml of 1 N H2 s o4 the unreacted H2 S o4 required 14.2 ml of 1 N n a o h for neutralization the equivalent weight of the metal is​
WILL BE MARKED AS BRAINLIEST IF CORRECT ANSWER

Answer 1

Explanation:

For the reaction between Sulphuric acid and sodium hydroxide ; let x ml of acid is reacted ; then according to normality equation : Acid Base

N1V1 = N2V2

1× x = 1 × 14.2 then x = 14.2 ml

Now the volume of acid reacts with Metal = 50 - 14.2 = 35.8 ml

Now for the reaction of acid and metal :

Number of gram equivalent of metal = number of gram equivalent of acid

mass of metal / equivalent weight = N V ( in L)

0.43 / E = 1 × 35.8 ×10-3 Then E = 12

Regards

Answer 2

$\huge\mathbb\green{Question}$

A metal weighing 0.43 gram was dissolved in 50 ml of 1 N H2 s o4 the unreacted H2 S o4 required 14.2 ml of 1 N n a o h for neutralization the equivalent weight of the metal is ?

WILL BE MARKED AS BRAINLIEST IF CORRECT ANSWER

$\huge\mathbf\red{required \: Answer}$

$Given$

weight of metal= 0.43g

it was dissolved in water=50ml

equivalant weight of metal=21.5

$\large\underline\purple{\sf hope \: it's \: helpful \: for \: you . }$

$\huge\mathbb\orange{MrRavi47}$