An aqueous solution of sodium stearate, C17H35 COONa, becomes a colloidal sol at the concentration of 1.2 × 10–3 M at 25°C. If at this concentration, 2.4 × 1013 colloidal (micelle) particles are present per mm3 , the number of stearate ions in each colloidal particle is : (NA = 6 × 1023)

1 mm3=10-9 m3 1 m3→ 103 litre 10-4 m3→ 10-6 litrenumber of particle=1.2×10-9×NA2.4×1013=12×10-22×6×1023=30

An aqueous solution of sodium stearate, C17H35 COONa, becomes a colloidal sol at the concentration of 1.2 × 10–3 M at 25°C. If at this concentration, 2.4 × 1013 colloidal (micelle) particles are present per mm3 , the number of stearate ions in each colloidal particle is : (NA = 6 × 1023)

1 mm3=10-9 m3 1 m3→ 103 litre 10-4 m3→ 10-6 litrenumber of particle=1.2×10-9×NA2.4×1013=12×10-22×6×1023=30

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

An aqueous solution of sodium stearate, $C_{17} H_{35} COONa$ , becomes a colloidal sol at the concentration of $1.2 \times 10 - 3 M at 25 ° C$ . If at this concentration, $2.4 \times 10^{13}$ colloidal (micelle) particles are present per ${mm}^{3}$ , the number of stearate ions in each colloidal particle is : $(NA = 6 \times 10^{23})$

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

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$1 {mm}^{3} = 10^{- 9} m^{3} 1 m^{3} \overset{}{\to} 10^{3} litre 10^{- 4} m^{3} \overset{}{\to} 10^{- 6} litre$

number of particle $= \frac{1.2 \times 10^{- 9} \times N_{A}}{2.4 \times 10^{13}}$

$= \frac{1}{2} \times 10^{- 22} \times 6 \times 10^{23} = 30$

Watch 3-min video & get full concept clarity

courses

No courses found

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Online Course for IIT JEE

Best Online Course for NEET

Best Online Course for CBSE

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Get Expert Academic Guidance – Connect with a Counselor Today!

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Online Course for IIT JEE

Best Online Course for NEET

Best Online Course for CBSE

best study material, now at your finger tips!

live classes
progress tracking
24x7 mentored guidance
study plan analysis

download the app

An aqueous solution of sodium stearate, C17H35 COONa, becomes a colloidal sol at the concentration of 1.2 × 10–3 M at 25°C. If at this concentration, 2.4 × 1013 colloidal (micelle) particles are present per mm3 , the number of stearate ions in each colloidal particle is : (NA = 6 × 1023)

Best Courses for You

Detailed Solution

courses

Get Expert Academic Guidance – Connect with a Counselor Today!

best study material, now at your finger tips!

download the app

Download the App