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The total number of ions present in 1ml of 0.1M barium nitrate Ba((NO3)2 solution is- (1)6.02×10*18 (2)6.02×10*19 (3)3×6.02×10*19 (4)3×6.02×10*18?
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The total number of ions present in 1ml of 0.1M barium nitrate Ba((NO3...
Solution:
Total number of ions present in 1ml of 0.1M barium nitrate Ba((NO3)2 solution can be calculated as follows:
Step 1: Write the balanced chemical equation for the dissociation of barium nitrate in water.
Ba((NO3)2) → Ba2+ + 2NO3-
Step 2: Calculate the number of moles of barium nitrate in 1ml of 0.1M solution.
0.1M = 0.1 moles/liter
1ml = 0.001 liters
Number of moles of barium nitrate = 0.1 x 0.001 = 0.0001 moles
Step 3: Calculate the number of ions produced by the dissociation of 0.0001 moles of barium nitrate.
1 mole of barium nitrate produces 1 mole of Ba2+ and 2 moles of NO3-
Hence, 0.0001 moles of barium nitrate will produce:
0.0001 moles x 1 ion/mole Ba2+ = 0.0001 ions of Ba2+
0.0001 moles x 2 ions/mole NO3- = 0.0002 ions of NO3-
Total number of ions = 0.0001 + 0.0002 = 0.0003 ions
Step 4: Convert the number of ions to scientific notation.
0.0003 = 3 x 10^-4
Therefore, the total number of ions present in 1ml of 0.1M barium nitrate Ba((NO3)2 solution is 3 x 10^18.
Answer: (3) 3 x 10^18
Total number of ions present in 1ml of 0.1M barium nitrate Ba((NO3)2 solution can be calculated as follows:
Step 1: Write the balanced chemical equation for the dissociation of barium nitrate in water.
Ba((NO3)2) → Ba2+ + 2NO3-
Step 2: Calculate the number of moles of barium nitrate in 1ml of 0.1M solution.
0.1M = 0.1 moles/liter
1ml = 0.001 liters
Number of moles of barium nitrate = 0.1 x 0.001 = 0.0001 moles
Step 3: Calculate the number of ions produced by the dissociation of 0.0001 moles of barium nitrate.
1 mole of barium nitrate produces 1 mole of Ba2+ and 2 moles of NO3-
Hence, 0.0001 moles of barium nitrate will produce:
0.0001 moles x 1 ion/mole Ba2+ = 0.0001 ions of Ba2+
0.0001 moles x 2 ions/mole NO3- = 0.0002 ions of NO3-
Total number of ions = 0.0001 + 0.0002 = 0.0003 ions
Step 4: Convert the number of ions to scientific notation.
0.0003 = 3 x 10^-4
Therefore, the total number of ions present in 1ml of 0.1M barium nitrate Ba((NO3)2 solution is 3 x 10^18.
Answer: (3) 3 x 10^18
Community Answer
The total number of ions present in 1ml of 0.1M barium nitrate Ba((NO3...
Molarity=moles/volume in litres
moles=molarity×volume
= 0.1×10-3
=10-4
Ba (NO3)2 - Ba2+ + 2NO-3
no of ions = moles of ions x Av. const
=3× 10-4× 6.02× 10 23
= 3× 6.02 ×10 19
moles=molarity×volume
= 0.1×10-3
=10-4
Ba (NO3)2 - Ba2+ + 2NO-3
no of ions = moles of ions x Av. const
=3× 10-4× 6.02× 10 23
= 3× 6.02 ×10 19
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The total number of ions present in 1ml of 0.1M barium nitrate Ba((NO3)2 solution is- (1)6.02×10*18 (2)6.02×10*19 (3)3×6.02×10*19 (4)3×6.02×10*18?
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The total number of ions present in 1ml of 0.1M barium nitrate Ba((NO3)2 solution is- (1)6.02×10*18 (2)6.02×10*19 (3)3×6.02×10*19 (4)3×6.02×10*18? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about The total number of ions present in 1ml of 0.1M barium nitrate Ba((NO3)2 solution is- (1)6.02×10*18 (2)6.02×10*19 (3)3×6.02×10*19 (4)3×6.02×10*18? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The total number of ions present in 1ml of 0.1M barium nitrate Ba((NO3)2 solution is- (1)6.02×10*18 (2)6.02×10*19 (3)3×6.02×10*19 (4)3×6.02×10*18?.
The total number of ions present in 1ml of 0.1M barium nitrate Ba((NO3)2 solution is- (1)6.02×10*18 (2)6.02×10*19 (3)3×6.02×10*19 (4)3×6.02×10*18? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about The total number of ions present in 1ml of 0.1M barium nitrate Ba((NO3)2 solution is- (1)6.02×10*18 (2)6.02×10*19 (3)3×6.02×10*19 (4)3×6.02×10*18? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The total number of ions present in 1ml of 0.1M barium nitrate Ba((NO3)2 solution is- (1)6.02×10*18 (2)6.02×10*19 (3)3×6.02×10*19 (4)3×6.02×10*18?.
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