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1.60 g of a metal were dissolved in HNO3 to prepare its nitate. The nitrate on strong heating gives 2 g of its oxide. The equivalent weight of metal is:    
  • a)
    16         
  • b)
    32        
  • c)
    48        
  • d)
    12
Correct answer is option 'B'. Can you explain this answer?
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1.60 g of a metal were dissolved in HNO3 to prepare its nitate. The ni...
Eq. of metal = Eq. of oxide
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1.60 g of a metal were dissolved in HNO3 to prepare its nitate. The ni...
Given:
- Mass of metal = 1.60 g
- Mass of oxide = 2 g

To find:
- Equivalent weight of the metal

Solution:

Step 1: Calculate the number of moles of the metal
The number of moles of the metal can be calculated using its mass and molar mass. The molar mass of the metal is not given in the question, so we need to calculate it.

We can assume that the metal forms a nitrate with a formula of M(NO3)2, where M represents the metal. The molar mass of M(NO3)2 can be calculated as follows:

Molar mass of M(NO3)2 = Molar mass of M + 2 * (Molar mass of NO3)
= M + 2 * (14 + 3 * 16)
= M + 2 * 62
= M + 124

Since 1.60 g of the metal reacts with HNO3 to form its nitrate, we can write the following equation:

M + HNO3 → M(NO3)2 + H2

From this equation, we can see that 1 mole of metal reacts with 2 moles of nitric acid to form 1 mole of M(NO3)2. Therefore, the number of moles of the metal is given by:

Number of moles of metal = Mass of metal / Molar mass of metal
= 1.60 g / (M + 124) g/mol
= 0.0129 mol / (M + 124)

Step 2: Calculate the number of moles of oxide
The number of moles of the oxide can be calculated using its mass and molar mass. The molar mass of the oxide is not given in the question, so we need to calculate it.

Since 2 g of the oxide is formed from the reaction of 1.60 g of the metal, we can write the following equation:

M(NO3)2 → M2O + 4NO2 + O2

From this equation, we can see that 1 mole of M(NO3)2 gives 1 mole of M2O. Therefore, the number of moles of the oxide is given by:

Number of moles of oxide = Mass of oxide / Molar mass of oxide
= 2 g / Molar mass of oxide
= 2 g / (M2O)

Step 3: Calculate the equivalent weight of the metal
The equivalent weight of an element is defined as the weight of the element that combines or displaces 1 g of hydrogen or 8 g of oxygen or 35.5 g of chlorine.

From the equation in Step 1, we can see that 0.0129 mol of the metal combines with 2 * 0.0129 mol of nitric acid to form 0.0129 mol of M(NO3)2.

From the equation in Step 2, we can see that 0.0129 mol of M(NO3)2 gives 0.0129 mol of M2O.

Therefore, the equivalent weight of the metal is given by:

Equivalent weight of metal = Mass of metal / Number of moles of metal
= 1.60 g /
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1.60 g of a metal were dissolved in HNO3 to prepare its nitate. The nitrate on strong heating gives 2 g of its oxide. The equivalent weight of metal is: a)16 b)32 c)48 d)12Correct answer is option 'B'. Can you explain this answer?
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1.60 g of a metal were dissolved in HNO3 to prepare its nitate. The nitrate on strong heating gives 2 g of its oxide. The equivalent weight of metal is: a)16 b)32 c)48 d)12Correct answer is option 'B'. Can you explain this answer? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about 1.60 g of a metal were dissolved in HNO3 to prepare its nitate. The nitrate on strong heating gives 2 g of its oxide. The equivalent weight of metal is: a)16 b)32 c)48 d)12Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 1.60 g of a metal were dissolved in HNO3 to prepare its nitate. The nitrate on strong heating gives 2 g of its oxide. The equivalent weight of metal is: a)16 b)32 c)48 d)12Correct answer is option 'B'. Can you explain this answer?.
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