Question

3.15 g oxalic acid (COOH), xH,0] was dissolved to make 500 ml solution on titration 33.36 ml of this N solution were neutralized by 50 ml NaOH.Calculate 15 value of x​

Answer 1

Answer:

Explanation:

Meq.of oxalic acid in 28mL=Meq.ofNaOH

=0.08×35

∴ Meq.of oxalic acid in 500mL

=50028×35×0.08=50

∴3.150m/2×1000=50⇒m=126

∴H2C2O4.xH2O=90+18x=126

or x=2

Answer 2

Given: Mass of oxalic acid dissolved $=3.15g$

The volume of the solution where oxalic acid is dissolved $=500ml$

To find: The value of $x$

Solution:

It is given mass of oxalic acid $=3.15g$

The volume of solution where oxalic acid is dissolved $=500ml$

The reaction is a neutralization reaction, therefore, the mili-equivalent of oxalic acid will be equal to mili-equivalent of NaOH.

∴ Mili-Equivalent of oxalic acid in $33.36 ml$ solution = Mili-equivalent of NaOH

The formula for mili-equivalent of NaOH is $=Normality*Volume$ $=0.08*50$

(Normality of NaOH is $0.08$)

∴ Mili-Equivalent of oxalic acid in one ml solution $=\frac{0.08*50}{33.36}$ ml

∴ Mili-Equivalent of oxalic acid in $500ml$ of solution $=\frac{0.08*50*500}{33.36} = 60$ (approx)

Now,

The Weight of oxalic acid given is $=3.15g$

Let the molecular weight of oxalic acid be $m$

To write the mili-equivalent weight in a different form,

∴ $\frac{3.15}{\frac{m}{2} }*1000=60$

On solving the equation,

⇒ $\frac{m}{2}=\frac{3.15*1000}{60}$

⇒ $m= \frac{3.15*1000*2}{60}$

⇒ $m=105$

The formula of oxalic acid as is given $=(COOH)_2.xH_2O$

The molecular weight found is $=105$

Now,

$H_2C_2O_4.xH_20=105$

⇒ $90+18x=105$

⇒ $18x=105-90$

⇒ $x=\frac{15}{18}$

⇒ $x=0.8$ ≅ 1 (nearly)

Final answer:

The value of $x$ is 1.