If 500 mL of CaCl2 solution contains 3.01×1022 chloride ions, molarity of the solution will be

CaCl2→Ca2++2Cl−Cl− in solution =3.01×1022 ionsNo. of CaCl2 present in solution =3.01×10222 =1.505×1022CaCl2Molarity of solution =No.of moles of CaCl2Vol.of solution (in litre)=1.51×10226.022×1023×1000500=0.05

If 500 mL of CaCl2 solution contains 3.01×1022 chloride ions, molarity of the solution will be

CaCl2→Ca2++2Cl−Cl− in solution =3.01×1022 ionsNo. of CaCl2 present in solution =3.01×10222 =1.505×1022CaCl2Molarity of solution =No.of moles of CaCl2Vol.of solution (in litre)=1.51×10226.022×1023×1000500=0.05

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

If 500 mL of $C a C l_{2}$ solution contains $3.01 \times 10^{22}$ chloride ions, molarity of the solution will be

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$0.1 M$

$0.02 M$

$0.01 M$

$0.05 M$

answer is A.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$C a C l_{2} \to C a^{2 +} + 2 C l^{-}$

$C l^{-}$ in solution $= 3.01 \times 10^{22}$ ions
No. of $C a C l_{2}$ present in solution $= \frac{3.01 \times 10^{22}}{2}$
$= 1.505 \times 10^{22} C a C l_{2}$
Molarity of solution $= \frac{N o . o f m o l e s o f C a C l_{2}}{V o l . o f s o l u t i o n (i n l i t r e)}$
$= \frac{1.51 \times 10^{22}}{6.022 \times 10^{23}} \times \frac{1000}{500} = 0.05$