Chemistry Solutions

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

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

1000 ml solution contains 0.02 milli mole (mm)

$\therefore$ 500 ml solution contains 0.02 m.m

$\therefore$ Solution made 1000 ml with H₂O

$\therefore$ m.m in final solution = 0.01 mm

Solution (A) + 0.01 m. m H₂SO₄

= 0.01 + 0.01

= 0.02 m.m

                             = 0.00002 * 10³ mm

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

$\Delta T_f=0.5°C$

K_f = 1.86

Using, density of water = 1g / mL

i = ?

$\Delta T_f=i{k}_f.m$           

$0.5=i*1.86*\frac{9.45}{94.5*500}*1000$          

i = 1.344

Now, using $\alpha =\frac{i-1}{n-1}$  

n for ClCH₂COOH = 2

 a =   $\frac{1.344-1}{2-1}=0.344$

Using

  $K_a=\frac{C{\alpha }^2}{1-\alpha }$           

$K_a=\frac{0.2*{\left(0.344\right)}^3}{0.66}=36*10^{-3}$           

           So; x = 36

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

$\pi =i*c*R*T=\frac{2*0.83*300}{60*10^3}=0.00415bar$

= 415 Pa

Using Raoult's Law, P_Total = P°_A·X_A + P°_B·X_B = (21 kPa * 1/3) + (18 kPa * 2/3) = 7 + 12 = 19 kPa.
Answer: 19 kPa

This is a limiting reactant problem. The initial moles are 5.25 mmol of Pb (NO? )? and 2.4 mmol of Cr? (SO? )? Based on the 3:1 stoichiometric ratio, Pb (NO? )? is the limiting reactant. The moles of PbSO? formed are equal to the initial moles of the limiting reactant, which is 5.25 mmol or 5.25 * 10? ³ moles. The question asks for the answer in units of 10? moles.

Using the Ideal Gas Law, PV = nRT:

P = 1 bar

V = 20 mL = 0.020 L

R = 0.083 L·bar·mol? ¹·K? ¹

T = 273 K

n = PV / RT = (1 * 0.020) / (0.0831 * 273) = 8.8 * 10? mol of Cl?

Number of Cl? molecules (N) = n * N_A = (8.8 * 10? ) * (6.022 * 10²³) = 5.3 * 10²? molecules.

Number of Cl atoms = 2 * (5.3 * 10²? ) = 1.06 * 10²¹.

The answer, rounded off to the nearest integer for the power of 10²¹, is 1.