Answer:

0.1M NaCl and0.1M glucose solution means that 0.1 moles of solute dissolve in 1L of solvent.

weight of solvent in both solution is same .
Dpression in freezing point can br calculated as :

$∆{\mathrm{T}}_{\mathrm{f}} = \frac{{\mathrm{K}}_{\mathrm{f}}\times {\mathrm{w}}_{\mathrm{B}}\times 1000}{{\mathrm{w}}_{\mathrm{A}} \times {\mathrm{M}}_{\mathrm{B}}}$

Where 
K_f = molal depression constant
w_B = weight of solute
w_A= weight of solvent
MB = molar mass of the solute

mass of NaCl = 58.69g/mol
mass of glucose = 180g/mol

The depression in freezingpoint is inversly proprotional to molar mass of solute thus more is molar mass of solute lesser is the depression in freezing point .
Thus it can explain that the NaCl wil be more than the glucose.

Answer:

by using formula .

$\frac{\mathrm{p}° -\mathrm{p}}{\mathrm{p}°} = x_{urea}$

let the mass of solution be 100g
therefore mass of urea = 10g
molecular mass of urea = 60 g

x_urea = 10/60 = 1/6

molecular mass of water = 18 g
 x_water = 90/18 = 5

mole fraction of urea =$$\begin{gathered} {\mathrm{x}}_{\mathrm{urea}} =\frac{1/6}{1/6+5} =\frac{ 1/6}{31/6} \\ {\mathrm{x}}_{\mathrm{urea}} =1/31 \\ \mathrm{using} \mathrm{above} \mathrm{formula} \\ \frac{55.3-\mathrm{p}}{55.3} =\frac{1}{31} \\ 55.3- \mathrm{p} =\frac{ 55.3}{31} \\ 55.3-\frac{ 55.3}{31} = \mathrm{p} \\ 55.3\left(1-\frac{1}{31}\right) =\mathrm{p} \\ \mathrm{p}= 55.3 \times \frac{30}{31} =53.52\mathrm{mm} \mathrm{Hg} \end{gathered}$$

Answer:

Copper sulphate conduct electricity only in molten state or in aqueous phase because on dissolving or melting it. copper sulphate dissociate in copper ion and sulphate ion and became free to move.
As they are freeto move in molten and aqueous phase the charge particle conduct electricity.

${\mathrm{CuSO}}_4 \to {\mathrm{Cu}}^{2+} + {\mathrm{SO}}_4^{2-}$

In solid and any other ionic form the ions held in a lattice and unable to move so they are not conduct electricity. 

Answer;

Let the mass of solution =100g

x_KOH = 30/56

volumeof solution = $\frac{\mathrm{mass} \mathrm{of} \mathrm{solution}}{\mathrm{density} \mathrm{of} \mathrm{solution}}$

$$\begin{gathered} = \frac{100\mathrm{g}}{\mathrm{d} \mathrm{g}/{\mathrm{cm}}^3} =\frac{ 100}{\mathrm{d}}\mathrm{cm}\times \frac{1}{1000} =\frac{1}{10\mathrm{d}}\mathrm{L} \\ \mathrm{Molarity} = \frac{\mathrm{number} \mathrm{of} \mathrm{moles} \mathrm{of} \mathrm{solute} }{\mathrm{volume} \mathrm{of} \mathrm{solution}} \\ 6.90 = \frac{30/56}{1/10\mathrm{d}} \\ 6.90 = \frac{30}{56}\times \frac{10\mathrm{d}}{1} \\ \mathrm{d} =1.288\mathrm{g}/\mathrm{cm} \end{gathered}$$

Answer:

When 30 ml of ethyl alcohol and 30 ml of water are mixed, the volume of resulting solution is more than 60ml.
It is because ethyl alchol and water do not form an ideal solution intermolecular forces of attraction are weak so the molecule will become farapart from each other hence volume will increase.