A detergent C12H25SO4−Na+ solution becomes a colloidal sol at a concentration of 10-3 M. On an average 1013 colloidal particles are present in 1 mm3. What is the average number of ions which are contained by one colloidal particle (micelle)? [Given: NA=6×1023mol−1]

No. of sodium lauryl sulphate moleculesCH3CH211SO4−Na+ in I litre solution =10−3×6×1023=6×1020No. of sodium lauryle sulphate molecules permm3= 6 x 1014No. of colloidal particles per mm3 = 1013No. of molecules per colloidal particle=6×10141013=60

A detergent C12H25SO4−Na+ solution becomes a colloidal sol at a concentration of 10-3 M. On an average 1013 colloidal particles are present in 1 mm3. What is the average number of ions which are contained by one colloidal particle (micelle)? [Given: NA=6×1023mol−1]

No. of sodium lauryl sulphate moleculesCH3CH211SO4−Na+ in I litre solution =10−3×6×1023=6×1020No. of sodium lauryle sulphate molecules permm3= 6 x 1014No. of colloidal particles per mm3 = 1013No. of molecules per colloidal particle=6×10141013=60

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A detergent $(C_{12} H_{25} {SO}_{4}^{-} {Na}^{+})$ solution becomes a colloidal sol at a concentration of 10^-3 M. On an average 10¹³ colloidal particles are present in 1 mm³. What is the average number of ions which are contained by one colloidal particle (micelle)? [Given: $N_{A} = 6 \times 10^{23} {mol}^{- 1}$ ]

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$6 \times 10^{7}$

None of these

answer is C.

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Detailed Solution

No. of sodium lauryl sulphate molecules
$({CH}_{3} {({CH}_{2})}_{11} {SO}_{4}^{-} {Na}^{+})$ in I litre solution
$\begin{array}{l} = 10^{- 3} \times 6 \times 10^{23} \\ = 6 \times 10^{20} \end{array}$
No. of sodium lauryle sulphate molecules per
mm³= 6 x 10¹⁴
No. of colloidal particles per mm³ = 10¹³
No. of molecules per colloidal particle
$= \frac{6 \times 10^{14}}{10^{13}} = 60$