102 - Physical Chemistry Questions Answers

mili eq of acid = VN = 20*0.1 = 2

mili eq of base = VN = 5*0.45 = 2.25

total milieq = 2.25-2 = 0.25 of base in 25 ml

so N = 0.25/25 = 10-2

so pOH = 2 so pH = 12

     A + 3B


--------->


2C + D

     1    1/2                0     0

    1-x   1/2-3x          2x    x

according to the given condition 1/2 - 3x = 2x

                                        so 1/10 = x

so 3x = 3/10

so % = 0.3*100/0.5 = 60

                      C₂H₄O -------------->           CO           +           CH₄

          t                                          P (C₂H₄O)

         0                                          116.6

         5                                          116.6-[122.56-116.6] = 116.6-5.96 = 110.64

         7                                           116.6-[125.72-116.6] = 116.6-9.12 = 107.48

   similarly others then find k for different readings by first order equation. k will be same for all readings.

because in ideal solution there is no interaction between the two liquids present in the solution

let x in HA and 1-x in HB

so according to the given condition -13.7x-12.7(1-x) = -13.5

solve

gas in gas is a solution but interaction between the molecules play almost no role so these solutions are addressed as mixtures

formula is πV = nRT or π = MRT,   M is molarity

strength of solution is 1.25m

so 1000 gm water ..............................................1.25 mole sucrose

so 1000 gm water .............................................1.25*342=427.5gm sucrose

so 1427.5 gm solution ....................................1.25 mole sucrose

so 1427.5/1.34=1065.30 ml = 1.0653 L solution ...........................1.25 mole sucrose

so M = 1.25/1.0653

put all the values to get answer.

CaCO₃     ---------->   CaO   +   CO₂ (g)

0.2 mole                   0            0

0.06                       0.14       0.14

so K_p= p                       only CO₂ is gas so its pressure is considered 

now p = 0.14*0.0821*1073/10 

now solve

 “weight of the dried protective agent in milligrams, which when added to 10 ml of a standard gold sol (0.0053 to 0.0058%) is just sufficient to prevent a colour change from red to blue on the addition of 1 ml of 10 % sodium chloride solution, is equal to the gold number of that protective colloid.”

0.25 g lyophilic colloid is added to 100 ml gold solution

so 250 mg lyophilic colloid is added to 100 ml gold solution

so 25 mg lyophilic colloid is added to 10 ml gold solution

so answer will be 25

its 4th option 

because As₂S₃ is usually a negative colloid and positive ion is responsible for its coagulation and according to Hardy Schulze priciple more positive ion will be more effective