Question

6.3 gram of crystalline oxalic acid is dissolved in water normality of solution is 0.01 N what is the volume of the solution​

Answer 1

Answer:

The volume of the solution​ is 10L.

Explanation:

First, let's see how normality is calculated. It is given as,

$\mathrm{N=\frac{w}{E\times V} }$         (1)

Where,

N=normality of the solution

w=weight of the substance

E=equivalent weight of the substance

V=volume of the solution

From the question, we have the following given values,

The weight of the oxalic acid(w)=6.3 gram

Normality of the solution(N)=0.01

Now let's calculate the equivalent weight of oxalic acid.

$\mathrm{E=\frac{M}{n} }$        (2)

M=molecular weight of oxalic acid=126gram

n=n-factor of oxalic acid=2

When the values are entered into equation (2), we obtain;

$\mathrm{E=\frac{126}{2}=63 }$          (3)

Equation (1) yields the following results when we insert the values:

$\mathrm{0.01=\frac{6.3}{63\times V} }$

$\mathrm{V=\frac{6.3}{63\times 0.01} }$

$\mathrm{V=10\ L} }$

Hence, the volume of the solution​ is 10L.

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