A soap solution contains 0.050 M soap with CMC = 1.0x10^-3 M and aggregation number 98. Take N_A = 6.022x10^23 mol^-1 and assume all soap above the CMC is micellised. Select the correct statements.

The micelle concentration is 5.0x10^-4 M

The hydrocarbon tails lie in the interior of each micelle and the ionic heads face the water

The number of micelles per litre is about 3.0x10^20

The micelle concentration is 5.0x10^-4 M

The hydrocarbon tails lie in the interior of each micelle and the ionic heads face the water

The number of micelles per litre is about 3.0x10^20

A 5.0x10^-4 M solution of the same soap would contain micelles of 98 anions

JEE Advanced levelOne or more correct optionsDifficulty 3 of 3sc156

A soap solution contains $0.050 M$ soap with CMC $= 1.0 \times 1 0^{- 3} M$ and aggregation number 98. Take $N_{A} = 6.022 \times 1 0^{23} mo l^{- 1}$ and assume all soap above the CMC is micellised. Select the correct statements.

AThe micelle concentration is $5.0 \times 1 0^{- 4} M$
BThe hydrocarbon tails lie in the interior of each micelle and the ionic heads face the water
CThe number of micelles per litre is about $3.0 \times 1 0^{20}$
DA $5.0 \times 1 0^{- 4} M$ solution of the same soap would contain micelles of 98 anions

Soap in micelles =

0.050 - 0.001 = 0.049

M, so micelle concentration =

0.049/98 = 5.0 \times 1 0^{- 4}

M. Number per litre =

5.0 \times 1 0^{- 4} \times 6.022 \times 1 0^{23} = 3.0 \times 1 0^{20}

. In water the tails are inside and the heads outside. A solution of

5.0 \times 1 0^{- 4}

M is below the CMC, so it contains no micelles.