Question

Constant current of 30 a is passed through an aqueous solution of nacl for a time of 1.0 hr. How many grams of naoh are produced? What is volume of cl, gas at s.T.P. Produced?

Answer 1

Answer :

The mass of NaOH produced are, 44.77 grams.

The volume of $Cl_2$ at STP is, 12.52 L

Explanation :

(1) First we have to calculate the mass of NaOH.

Using Faraday first law,

$w=\frac{E\times I\times t}{F}$

where,

w = mass

E = equivalent weight

I = current = 30 A

t = time = 1 hr = 3600 s

F = Faraday constant = 96500

Now we have to calculate the equivalent weight of NaOH.

$\text{Equivalent weight of }NaOH=\frac{\text{Molar mass of }NaOH}{n}$

where,

n = valency factor = 1  (For NaOH)

$\text{Equivalent weight of }NaOH=\frac{40}{1}=40$

Now put all the given values in the above formula, we get:

$w=\frac{E\times I\times t}{F}$

$w=\frac{40\times 30\times 3600}{96500}=44.77g$

The mass of NaOH produced are, 44.77 grams.

(2) Now we have to calculate the mass of $Cl_2$.

Using Faraday first law,

$w=\frac{E\times I\times t}{F}$

First we have to calculate the equivalent weight of $Cl_2$.

$\text{Equivalent weight of }Cl_2=\frac{\text{Molar mass of }Cl_2}{n}$

where,

n = valency factor = 2

$\text{Equivalent weight of }Cl_2=\frac{71}{2}=35.5$

Now put all the given values in the above formula, we get:

$w=\frac{E\times I\times t}{F}$

$w=\frac{35.5\times 30\times 3600}{96500}=39.7g$

Now we have to calculate the moles of $Cl_2$.

$\text{Moles of }Cl_2=\frac{\text{Mass of }Cl_2}{\text{Molar mass of }Cl_2}=\frac{39.7g}{71g/mol}=0.559mol$

Now we have to calculate the volume of $Cl_2$ at STP.

As, 1 mole of $Cl_2$ gas contains 22.4 L volume of $Cl_2$ gas

So, 0.559 mole of $Cl_2$ gas contains 0.559 × 22.4 = 12.52 L volume of $Cl_2$ gas

The volume of $Cl_2$ at STP is, 12.52 L