Calculate molarity of cacl2 solution on chemical analysis it is found that 500 ml of cacl2 solution contains 1.505 into 10 raised to 23 CL Negative Ion
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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
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