When dT = Kf * molality * i
= Kf*m*i
and when molality = (no of moles of solute) / Kg of solvent
= 2.5g /250g x 1 mol /85 g x1000g/kg
=0.1176 molal
and Kf for water = - 1.86 and dT = -0.255
by substitution
0.255 = 1.86* 0.1176 * i
∴ i = 1.166
when the degree of dissociation formula is: when n=2 and i = 1.166
a= i-1/n-1 = (1.166-1)/(2-1) = 0.359 by substitution by a and c(molality) in K formula
∴K = Ca^2/(1-a)
= (0.1176 * 0.359)^2 / (1-0.359)
= 2.8x10^-3
570/8.5=67.0 58... you only have to take the natural part, si the answer is 67 students
10g
Explanation:
Box 1, Mass of A = 10g
Box 2, Mass of B = 5g
Box 3, = 1A + 1B
Unknown:
Mass of B that would combine with mass of 20g of A
Solution:
Mass ratio of A to B:
= mass ratio
= mass ratio
The mass ratio of A to B = 2: 1
Now, number of B that will combine with 20g of A;
= mass ratio
= 
Mass of B = 10g
10g of B would combine with 20g of A
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Answer:
Mass of original sample = 100 g
Explanation:
Half life of cesium-137 = 30.17 years
Where, k is rate constant
So,
The rate constant, k = 0.02297 year⁻¹
Time = 90.6 years
Using integrated rate law for first order kinetics as:
Where,
is the concentration at time t
is the initial concentration
Initial concentration
= ?
Final concentration
= 12.5 grams
Applying in the above equation, we get that:-
![[A_0]=\frac{12.5}{e^{-0.02297\times 90.6}}\ g=100\ g](https://tex.z-dn.net/?f=%5BA_0%5D%3D%5Cfrac%7B12.5%7D%7Be%5E%7B-0.02297%5Ctimes%2090.6%7D%7D%5C%20g%3D100%5C%20g)
<u>Mass of original sample = 100 g</u>