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X and H form two compounds. They contain 17.65% and 12.50% hydrogen by mass respectively. If the first compound is XH3 then second one could be options: 1)X2H4 2)XH 3)X2H 4)X3H?
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X and H form two compounds. They contain 17.65% and 12.50% hydrogen by...
Let m.....be atomic mass of ..Xin first compound....XH3%by mass of hydrogen = mass of hydrogen/ molar mass=>17.65/100 = 3 /(m+3) =>17.65×(3+m) = 300 =>17.65m = 300 - 17.65×3 = 300 - 52.95 =>17.65m = 247.05 => m = 247.05/17.65 ≈ 14gin second compound ......XH(y).... %by mass of hydrogen = mass of hydrogen/ molar mass=> 12.5/100 = y/(14+y)=> 12.5×(14+y) = 100y=> 12.5×14 = (100-12.5)y=> y = 175/87.5= 2it means .....second compound is....XH2....but as in option... it is not given......The answer would be......X2H4 [ as X2H4 and XH2 ...have ...X. & H in same ratio....which is ...{1:2}.....]
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X and H form two compounds. They contain 17.65% and 12.50% hydrogen by...
Understanding the Mass Percentages
To determine the formula of the second compound based on the given mass percentages of hydrogen, we first analyze the first compound, XH3.
Calculating Molar Mass of XH3
- The percentage of hydrogen in XH3 is 17.65%.
- The molar mass of hydrogen (H) is approximately 1 g/mol.
- For XH3, the mass of hydrogen is 3 g/mol, while the total molar mass (M) can be calculated as follows:
17.65% = (3 g/mol / M) × 100
- Rearranging gives M = (3 g/mol / 17.65) × 100 ≈ 17.0 g/mol.
- Therefore, the mass of X in XH3 is approximately 14.0 g/mol (17.0 g/mol - 3 g/mol).
Analyzing Compound H
Now, let’s analyze the second compound, which has 12.50% hydrogen by mass.
Calculating Molar Mass of the Second Compound
- For the second compound, we use the same logic:
12.50% = (mass of H / total molar mass) × 100.
- Assuming the second compound is of the form XnHm, we can express the mass of hydrogen (H) as m g/mol.
- Rearranging gives total molar mass (M) = (m g/mol / 12.50) × 100.
Evaluating the Options
Now, let’s evaluate the possible compounds:
1. X2H4:
- Molar mass = 2(14) + 4(1) = 28 + 4 = 32 g/mol.
- H% = (4 / 32) × 100 = 12.5% (fits).
2. XH:
- Molar mass = 14 + 1 = 15 g/mol.
- H% = (1 / 15) × 100 = 6.67% (doesn't fit).
3. X2H:
- Molar mass = 2(14) + 1 = 29 g/mol.
- H% = (1 / 29) × 100 = 3.45% (doesn't fit).
4. X3H:
- Molar mass = 3(14) + 1 = 43 g/mol.
- H% = (1 / 43) × 100 = 2.33% (doesn't fit).
Conclusion
The only suitable compound that matches the 12.50% hydrogen requirement is X2H4.
To determine the formula of the second compound based on the given mass percentages of hydrogen, we first analyze the first compound, XH3.
Calculating Molar Mass of XH3
- The percentage of hydrogen in XH3 is 17.65%.
- The molar mass of hydrogen (H) is approximately 1 g/mol.
- For XH3, the mass of hydrogen is 3 g/mol, while the total molar mass (M) can be calculated as follows:
17.65% = (3 g/mol / M) × 100
- Rearranging gives M = (3 g/mol / 17.65) × 100 ≈ 17.0 g/mol.
- Therefore, the mass of X in XH3 is approximately 14.0 g/mol (17.0 g/mol - 3 g/mol).
Analyzing Compound H
Now, let’s analyze the second compound, which has 12.50% hydrogen by mass.
Calculating Molar Mass of the Second Compound
- For the second compound, we use the same logic:
12.50% = (mass of H / total molar mass) × 100.
- Assuming the second compound is of the form XnHm, we can express the mass of hydrogen (H) as m g/mol.
- Rearranging gives total molar mass (M) = (m g/mol / 12.50) × 100.
Evaluating the Options
Now, let’s evaluate the possible compounds:
1. X2H4:
- Molar mass = 2(14) + 4(1) = 28 + 4 = 32 g/mol.
- H% = (4 / 32) × 100 = 12.5% (fits).
2. XH:
- Molar mass = 14 + 1 = 15 g/mol.
- H% = (1 / 15) × 100 = 6.67% (doesn't fit).
3. X2H:
- Molar mass = 2(14) + 1 = 29 g/mol.
- H% = (1 / 29) × 100 = 3.45% (doesn't fit).
4. X3H:
- Molar mass = 3(14) + 1 = 43 g/mol.
- H% = (1 / 43) × 100 = 2.33% (doesn't fit).
Conclusion
The only suitable compound that matches the 12.50% hydrogen requirement is X2H4.
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X and H form two compounds. They contain 17.65% and 12.50% hydrogen by mass respectively. If the first compound is XH3 then second one could be options: 1)X2H4 2)XH 3)X2H 4)X3H?
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X and H form two compounds. They contain 17.65% and 12.50% hydrogen by mass respectively. If the first compound is XH3 then second one could be options: 1)X2H4 2)XH 3)X2H 4)X3H? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about X and H form two compounds. They contain 17.65% and 12.50% hydrogen by mass respectively. If the first compound is XH3 then second one could be options: 1)X2H4 2)XH 3)X2H 4)X3H? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for X and H form two compounds. They contain 17.65% and 12.50% hydrogen by mass respectively. If the first compound is XH3 then second one could be options: 1)X2H4 2)XH 3)X2H 4)X3H?.
X and H form two compounds. They contain 17.65% and 12.50% hydrogen by mass respectively. If the first compound is XH3 then second one could be options: 1)X2H4 2)XH 3)X2H 4)X3H? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about X and H form two compounds. They contain 17.65% and 12.50% hydrogen by mass respectively. If the first compound is XH3 then second one could be options: 1)X2H4 2)XH 3)X2H 4)X3H? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for X and H form two compounds. They contain 17.65% and 12.50% hydrogen by mass respectively. If the first compound is XH3 then second one could be options: 1)X2H4 2)XH 3)X2H 4)X3H?.
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