Question

25 ml of hcl librates 10 ml of CO2 at ntp when reacting with excess of caco3 the normality of hcl be

Answer 1

Answer:

(ii) Volume strength of H O = 5.6 x Normality of H O ... The molarity of a solution obtained by mixing 750 mL of 0.5 M HCl and 250 ... treating 25 g of this sample with excess of dil HCI.

Answer 2

The Normality Of HCl Is 0.03 N

GIVEN

The volume of HCl = 25 ml

the volume of CO₂ formed = 10 ml

TO FIND

Normality of HCl

SOLUTION

We can solve the above problem as follows -

We know that,

Normality = Molarity × n-factor

Where,

n-factor = number of replaceable hydrogen ions ( in case of an acid)

To find the normality of HCl, let us first write the chemical formula and find the molarity of HCl.

CaCO₃ + 2HCl ------------> CaCl₂ + CO₂ + H₂O

The volume of Carbon dioxide gas formed = 10 ml at Ntp

10 mL = 0.01L

we know,

1 mole of gas occupies 22.4 L at Ntp

so,

$0.01L = \frac{1}{22.4} \times 0.01 = 4.4 \times {10}^{ - 4} moles$

Now, From the above reaction, we can say that -

Moles of HCl = 2 × moles of CO₂

so,

Moles of HCl = 2× 4.4×10⁻⁴ = 8.8 × 10⁻⁴ moles.

Molarity = Moles of solute /volume of solute in 1 liter

Therefore,

$Molarity \: of \: HCl = \frac{8.8 \times {10}^{ - 4} \times {10}^{3} }{25}$

= 0.03 M

Now,

We know that the n-factor of HCl is 1

So, Normality = Molarity = 0.03

Hence, the Normality of HCl is 0.03 N

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