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
I’m not your just blind. I’m sorry
Answer:
See below in bold.
Explanation:
a) Yes because it is an ionic compound. Forms ions in water solution.
b) This is an alcohol with covalent bonds it does not ionise in water - no.
c) Yes this forms ions in water solution.
d) This is a sugar and does not ionise so No.
Answer: The final temperature (T) will be;
16.23°c
Explanation: To find the final temperature of any mixture of substance with an initial temperature use the formula;
M1c(T-T1) + M2c(T-T2)=0
c is the specific heat capacity of the two different substance, but because ice and water are the same, we assume c to be 1
T= final temperature of the mixture
T1= initial temperature of the ice= -10.2°c
T2= initial temperature of the water= 19.7°c
M1= mass of ice= 33.1g
M2= mass of water= 251g
Using the formula above
33.1(T-(-10.2)) + 251(T-19.7)=0
Solving out the bracket
33.1T + 337.62 + 251T - 4944.7 = 0
Collecting like terms to both side of the equation and solving
33.1T + 251T = 4944.7 - 33.62
284.1T = 4607.08
T = 4607.08÷284.1 = 16.23°c
Answer:
0.091 mol/L
Explanation:
Molarity of a substance , is the number of moles present in a liter of solution .
M = n / V
M = molarity ( unit = mol / L or M )
V = volume of solution in liter ( unit = L ),
n = moles of solute ( unit = mol ),
Moles is denoted by given mass divided by the molecular mass ,
Hence ,
n = w / m
n = moles ,
w = given mass ,
m = molecular mass .
From the data of the question , it is given that ,
w = 1.5 g
V = 220 mL
Since ,
1 mL = 1/1000L
V = 0.220 L
As we known , the mass of bleach NaOCl is ,
m = 74.44 g/mol
from the above equation , moles can be calculated as -
n = w / m
n = 1.5g / 74.44 g/mol
n = 0.0201 mol
Molarity of the solution is calculated as ,
M = n / V
M = 0.0201 mol / 0.220 L
M = 0.091 mol/L