Question

1.26 g of hydrated oxalic acid was dissolved in water to prepare 250 ml of solution . calculate molarity of solution

Answer 1

Equivalent weight of oxalic acid is 63 gram

We are using 1.26 gram

Volume of the solution is 250 ml

Now,

Normality of oxalic acid=

(weight of acid×1000)/(eq. Weight×vol.of sol.)

= (1.26×1000)/(63×250)

=0.08 N

Normality of given solution is 0.08 normal.

Answer 2

✳️✳️✳️Equivalent weight of oxalic acid is 63 gram

We are using 1.26 gram

Volume of the solution is 250 ml

Now,

Normality of oxalic acid=

(weight of acid×1000)/(eq. Weight×vol.of sol.)

= (1.26×1000)/(63×250)

= $\bold{\huge{\boxed{\red{0.08 N}}}}$

Normality of given solution is $\bold{\huge{\underline{\underline{\red{0.08 \: Normal}}}}}$.✳️✳️✳️

MAY THIS HELP U MY FRND!!☺️✌️♥️♥️