Solutions

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

i = total no. of particles after dissociation/association / total no. of particle before dissociation/association
HA? H? + A? (Dissociation)
0.5a
2HA → (HA)? (Association)
0.5a/2
i = (a + 0.25a)/a = 125 x 10? ²
Ans. = 125

Molality = moles of solute / mass of solvent (kgs) ⇒ 1 = 0.5 / W (kgs)
W_solvent (kgs) = 0.5 = 500 g

m = 10 molal
K_b = 0.5 K kg mol? ¹
Using: ΔT_b = I K_b m
and α = (i - 1) / (n - 1)
n for AB? is 3; α = 0.1
0.1 = (i - 1) / (3 - 1) ⇒ I = 1.2
ΔT_b = 1.2 * 0.5 * 10 = 6 °C
So, boiling point of solution = 100 + 6 = 106 °C

A 6.5 molal solution means 6.5 moles of KOH is in 1 kg (1000 g) of solvent (H? O).
Moles of solute, n_B = 6.5
Mass of solute, W_B = 6.5 * 56 = 364 g
Mass of solvent, W_A = 1000 g
Mass of solution = 1364 g
Volume of solution = 1364 / 1.89 mL
Now, molarity = [6.5 / (1364 / 1.89)] * 1000 M = 9 M