Same amount of electric current is passed through solutions of AgNO₃ and HCI. If 1.08 g of silver is obtained in the first case, the volume of hydrogen liberated at S.T.P. in the second case isa)22400 mlb)2240 mlc)224 mld)112 mlCorrect answer is option 'D'. Can you explain this answer?

Question

Same amount of electric current is passed through solutions of AgNO&#8323; and HCI. If 1.08 g of silver is obtained in the first case, the volume of hydrogen liberated at S.T.P. in the second case isa)22400 mlb)2240 mlc)224 mld)112 mlCorrect answer is option 'D'. Can you explain this answer?

Answer

Number of equivalent for Ag =( 1.08/108 )=0.01eq

Since 1 eq. produces 11.2 litre of H2 at STP

So 0.01 eq. produces (11.2�0.01) litre of H2

Or vol=0.112 litre

=112ml

Answer

Given:
Mass of silver obtained in the first case = 1.08 g
Volume of hydrogen liberated in the second case = ?

We can use Faraday's laws of electrolysis to solve this problem.

1st law of electrolysis:
The amount of substance produced by an electric current during electrolysis is directly proportional to the quantity of electric charge passed through the electrolyte.

2nd law of electrolysis:
The amounts of different substances produced by the same quantity of electricity passing through different electrolytes are proportional to their chemical equivalent weights.

From the given information,
Amount of silver deposited = 1.08 g
Chemical equivalent weight of silver = atomic weight / valency = 108 / 1 = 108 g/equivalent
So, the number of equivalents of silver deposited = 1.08 / 108 = 0.01 equivalents

Since the same amount of electric current is passed through the solution of HCl, the number of equivalents of hydrogen gas liberated will be the same as the number of equivalents of silver deposited in the first case.

Chemical equivalent weight of hydrogen = 1 g/equivalent
So, the mass of hydrogen liberated = number of equivalents of hydrogen x chemical equivalent weight of hydrogen
= 0.01 x 1 = 0.01 g

At STP, the volume of 1 g of hydrogen gas is 22.4 L or 22400 mL.
Therefore, the volume of 0.01 g of hydrogen gas is 0.01 x 22400 = 224 mL or 0.224 L

Hence, the correct answer is option (d) 112 mL (as the question asks for the volume of hydrogen gas liberated and not the volume of 1 g of hydrogen gas).

Answer

Solution:

We know that the same amount of electric current is passed through solutions of AgNO3 and HCl.

Hence, the number of electrons passing through each solution is the same.

Now, let's calculate the number of moles of silver deposited in the first case.

The molar mass of silver is 107.87 g/mol.

The mass of silver obtained is 1.08 g.

Number of moles of silver = 1.08/107.87 = 0.01 mol

Now, let's calculate the volume of hydrogen liberated in the second case.

From the balanced chemical equation:

2HCl + 2e- → H2 + 2Cl-

1 mole of electrons liberates 1 mole of hydrogen gas.

1 mole of H2 at STP occupies 22.4 L.

Number of moles of electrons passed = 0.01 mol (from the first case)

Number of moles of hydrogen liberated = 0.01 mol

Volume of hydrogen gas liberated at STP = 0.01 mol x 22.4 L/mol = 0.224 L = 224 mL

Hence, the correct option is (d) 112 mL.

Answer

Solution:
The balanced chemical equation for the electrolysis of AgNO3 and HCl is given below:

- Electrolysis of AgNO3: 2AgNO3 → 2Ag + O2 + 2NO2
- Electrolysis of HCl: 2HCl → H2 + Cl2

Given:
- Mass of silver obtained from AgNO3 solution = 1.08 g
- Same amount of current is passed through both solutions

To find: Volume of hydrogen liberated from HCl solution at STP

Step 1: Calculate the number of moles of Ag obtained from AgNO3 solution
Molar mass of Ag = 107.87 g/mol
Number of moles of Ag = Mass of Ag / Molar mass of Ag
Number of moles of Ag = 1.08 g / 107.87 g/mol
Number of moles of Ag = 0.01 mol

Step 2: Use stoichiometry to find the number of moles of H2 produced from HCl solution
From the balanced chemical equation, we know that 2 moles of HCl produce 1 mole of H2
Number of moles of H2 = (0.01 mol Ag / 2 mol AgNO3) x (2 mol HCl / 2 mol Ag) x (1 mol H2 / 2 mol HCl)
Number of moles of H2 = 0.0025 mol

Step 3: Use the ideal gas law to find the volume of H2 at STP
PV = nRT, where P = 1 atm, V = volume in liters, n = number of moles, R = 0.0821 L.atm/mol.K, T = 273 K
V = nRT / P
V = 0.0025 mol x 0.0821 L.atm/mol.K x 273 K / 1 atm
V = 0.056 L or 56 ml

Step 4: Convert the volume to ml
1 L = 1000 ml
0.056 L = 56 ml

Therefore, the volume of hydrogen liberated at STP from the HCl solution is 56 ml, which is closest to option D (112 ml).

Answer

Number of equivalent for Ag =(1081.08)=0.01eq

Since 1 eq. produces 11.2 litre of H2 at STP

So 0.01 eq. produces (11.2×0.01) litre of H2

Or vol=0.112 litre

=112cm3

Answer

AgNO3 breaks as Ag+ and NO3- means...it requires 1 mole of electron to deposit 108 g of Ag ..i.e.,96500C charge ....so the charge required to deposite 1.08 g is 965C charge.....
HCl breaks as H+and Cl- ....and 2 H+→H2....means 2 moles of electron liberates 22400 ml H2....i.e.,193000C charge liberates 22400 ml H2.......so,965 C charge will liberate H2=22400*965/193000=112ml

Answer

Mass of AgNo3\mass of H2 = Eq mass of AgNo3/ Eq mass . of H2
1.08/m=108/1

m= 1.08/108
m=0.01 = 10–²
2g of H2 at stp occupies 22400 cm³
10–² g of H2 occupies at stp
22400/2 ×10–²×10–² cm³ = 11200×10–4
= 112cm³

Answer

2g of H₂ at S.T.P. occupy 22400 cm³
∴10⁻² g of H₂at S.T.P. occupy

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