Calculate the molarity of a solution of cacl2. if on chemical analysis it is found that 200 ml of CaCl2 contains 3.01×10²² Cl- ions
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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
hope it will help you
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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
｡^‿^｡｡^‿^｡｡^‿^｡
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