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Using electrolytic method, if cost of pr...

Using electrolytic method, if cost of production of 10L of oxygen at STP is Rs. x, the cost of production of same volume of hydrogen at STP will be:

A

2x

B

x/16

C

x/32

D

x/2

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The correct Answer is:
To solve the problem of determining the cost of production of 10L of hydrogen at STP, given that the cost of production of 10L of oxygen at STP is Rs. x, we can follow these steps: ### Step-by-Step Solution: 1. **Understand the Electrolysis of Water**: - The electrolysis of water produces hydrogen and oxygen in a specific volume ratio. The reaction can be represented as: \[ 2 H_2O \rightarrow 2 H_2 + O_2 \] - From this reaction, we can see that for every 2 volumes of hydrogen produced, 1 volume of oxygen is produced. This establishes a volume ratio of hydrogen to oxygen as 2:1. 2. **Given Information**: - We know that the cost of producing 10L of oxygen is Rs. x. 3. **Determine the Volume of Hydrogen**: - Since the volume ratio of hydrogen to oxygen is 2:1, if we have 10L of oxygen, the corresponding volume of hydrogen produced would be: \[ \text{Volume of Hydrogen} = 2 \times \text{Volume of Oxygen} = 2 \times 10L = 20L \] 4. **Cost of Hydrogen Production**: - The cost of production is proportional to the volume produced. If the cost of producing 10L of oxygen is Rs. x, then the cost of producing 20L of hydrogen can be calculated as follows: \[ \text{Cost of 20L Hydrogen} = \text{Cost of 10L Oxygen} \times 2 = x \times 2 = 2x \] 5. **Final Answer**: - Therefore, the cost of production of 10L of hydrogen at STP is Rs. 2x. ### Summary: - The cost of producing 10L of hydrogen at STP is Rs. 2x.

To solve the problem of determining the cost of production of 10L of hydrogen at STP, given that the cost of production of 10L of oxygen at STP is Rs. x, we can follow these steps: ### Step-by-Step Solution: 1. **Understand the Electrolysis of Water**: - The electrolysis of water produces hydrogen and oxygen in a specific volume ratio. The reaction can be represented as: \[ 2 H_2O \rightarrow 2 H_2 + O_2 ...
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