Answer:



Among H,Cl and F, Hydrogen is least electronegtive while F is most electronegtive than Cl and H.
Thus F can withdraws more electron towards itslef more than Cl and H. So trifluoroacetic acid can easily lose the H⁺ ions. i.e.
trifluroacetic acid ionize to the larger extent .
Now more the ion produces the greater is the dpress ion of the freezing point 
Hence, the depression of freezing point increase in order :

Acetic acid<trichloracetic acid < trifluroacetic acid

Answer:

Mol. mass of benzoic acid, C₆H₅COOH
= 6 x 12 + 5 x 1 + 12 + 16 + 16 + 1
= 72 + 5 + 12 + 16+16 + 1
= 122 g mol^–1

by using formula;

$M =\frac{ x}{molecular mass of given subs\tan ce}\times \frac{1000}{required volume}$

here x= amount of substance required 

$0.15M =\frac{ x}{122}\times \frac{1000}{250}$

Amount of benzoic acid required

$$\begin{gathered} =\frac{122}{1000}\times 250\times 0.15 \\ = \frac{4575}{1000} = 4.575 \mathrm{g}. \end{gathered}$$

Answer:

number of moles present in 92g of Na⁺ ion.
         =92/23g mol^-1 = 4mol

Molality, $\mathrm{m}=\frac{\mathrm{No}. \mathrm{of} \mathrm{gm} \mathrm{moles} \mathrm{of} \mathrm{solute}}{\mathrm{wt}. \mathrm{of} \mathrm{solvent} \mathrm{in} \mathrm{kg}}$
No. of gm moles of solute = 92/23 = 4
Wt. of water (solvent) = 1 kg
$\mathrm{Molality} = \frac{4}{1} = 4 \mathrm{m}.$

Answer:

Mass of solute  = 6.5 g
Mass of solution = 450 + 6.5 = 456.6 g

$$\begin{gathered} \mathrm{Mass} \mathrm{percentage} = \frac{\mathrm{Mass} \mathrm{of} \mathrm{solute}}{\mathrm{Mass} \mathrm{of} \mathrm{solution}}\times 100 \\ = \frac{6.5}{456.5}\times 100 = \frac{650}{456.5} = 1.424\%. \end{gathered}$$