Solutions

Let we take $1\mathcal{l}$  of solution

 Mass of solute = Volume * Density

= 0.5 $m\mathcal{l}*1.05gm/m\mathcal{l}$  

= 0.525 gram

Mass of solution = 1 kg. [considering very dilute solution]

Mass of solvent = 1000 – 0.525 = 999.475 gram

$\Rightarrow i=1.9$  

$i=\frac{m_{theor}}{m_{app}}=\frac{30}{m_{app}}$

$\Delta T_b=i{k}_b.m]\underset{m-x}{\overset{m}{HA}}=\underset{x}{\overset{-}{H^{+}}}+\underset{x}{\overset{-}{A^{-}}}$

$0.0156=\left(\frac{m*x}{m}\right)*0.52*m$

m + x = 0.03

$\Rightarrow x=0.01\to i=1.5=\frac{30}{m_{app}}\Rightarrow m_{appp}=20$

At pH = 6.4

As in case of H₂CO₃,

pH = pKa it will be only when

[weak acid] = [conjugate base].

In case of 2

H₂PO^-₄ |HPO^2-₄

Solute and solvent both should obey Raoult's law

Molarity $=\frac{\text{moles of}\text{solute}}{\text{volume of solution in}L}=\frac{1*1.15}{1342}*1000=0.86\mathrm{M}$

∴ % of S in the compound
= (32/233) * (mass of BaSO? / mass of compound) * 100 = (32 * 0.35 * 100) / (233 * 0.25) = 19.227 ≈ 19.23

Let V ml is taken

  $\frac{5*V+1*150}{\left(V+150\right)+0.9}=2.5$            

5V + 150 = 2.25 V + 337.5

$V=\frac{187.5}{2.75}=68.18ml$            

? Solvent is H? O, which is in excess

So using m ( molality ) = (x? *1000)/ (x? * (M? )? )

? x? = 0.74 (Mol? = 18 g)

x? = 1 – 0.74 = 0.26 ∴ m = (0.26 * 1000)/ (0.74 * 18) = 19.5

W = 4.5 g

M.W = 90

V = 250 ml

Using M = $\frac{W}{M.W*V}*1000$  

$M=\frac{4.5}{90*250}*1000=0.2$       

M = 2 * 10^-1 M

So, x = 2

ppm of O? = (wt. of O? ) / (wt. of H? O) * 10?
= (10.3 mg) / (1.03 * 10? mg) * 10?
= 10ppm