New data must be found to support it
Explanation:
The given data is as follows.
q = 6.0 nC = 
inner radius (r) = 1.0 cm = 0.01 m (as 1 cm = 100 m)
So, there will be same charge on the inner surface as the charge enclosed with an opposite sign.
Formula to calculate the charge density is as follows.
.......... (1)
Since, area of the sphere is as follows.
A =
........... (2)
Hence, substituting equation (2) in equation (1) as follows.

=
= 
or, = 4.77 
Thus, we can conclude that the resulting charge density on the inner surface of the conducting sphere is 4.77
.
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Answer
is: V<span>an't
Hoff factor (i) for this solution is 1,81.
Change in freezing point from pure solvent to
solution: ΔT =i · Kf · b.
Kf - molal freezing-point depression constant for water is 1,86°C/m.
b - molality, moles of solute per
kilogram of solvent.
</span><span>b = 0,89 m.
ΔT = 3°C = 3 K.
i = </span>3°C ÷ (1,86 °C/m · 0,89 m).
i = 1,81.