Calculate molarity of solution of cacl2 if on chemical analysis it is found that 200 ml of cacl2 contains 3.01 into 10 to the power 22 cl negative ions
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Answered by
75
Since CaCl2 has 2 chloride ions, the number of CaCl2 “molecules” in 200mL is
= 3.01x10^22 ÷ 2
= 3.01x10^22 ÷ 2 x 5 per litre
= 7.525 x 10^22 “molecules”
1 mole of a substance contains 6.02 x 10^23 “molecules”
Therefore, number of “molecules” of CaCl2 / litre
= 7.525 x 10^22 ÷ 6.02 x 10^23
= 0.125
CaCl2 solution is 0.125 M
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= 3.01x10^22 ÷ 2
= 3.01x10^22 ÷ 2 x 5 per litre
= 7.525 x 10^22 “molecules”
1 mole of a substance contains 6.02 x 10^23 “molecules”
Therefore, number of “molecules” of CaCl2 / litre
= 7.525 x 10^22 ÷ 6.02 x 10^23
= 0.125
CaCl2 solution is 0.125 M
HOPE IT WILL HELP YOU 。^‿^。。^‿^。
Answered by
16
Answer: 0.125M is the answer
Explanation:
CaCl2 has 2 chloride ions = 3.01 X 10^22/2
converting 200ml = 1/5L
=3.01 X 10^22/2 X 5
=7.5 X 10^22 molecules
now, number of molecules of CaCl2
= 7.525 X 10^22/6.02X10^23 (as, 1 mole contains 6.023X10^23 molecules)
= 0.125M
Molarity of CaCl2 solution is 0.125M
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