Question

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

Answer 1

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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Answer 2

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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