Answer: Option (A) is the correct answer.
Explanation:
Rate of diffusion is defined as the total movement of molecules from a region of higher concentration to lower concentration.
The interaction between medium and the material is responsible for the rate of diffusion of a material or substance.
A small concentration gradient means small difference in the number of molecules taking part in a reaction. So, when there no large difference between the concentration then there won't be much difference in the rate of diffusion of a material.
Whereas a higher concentration of molecules will lead to more number of collisions due to which frequency of molecules increases. Therefore, rate of diffusion will also increase.
Small molecule size will also lead to increases in rate of diffusion. This is because according to Graham's law rate of diffusion is inversely proportional to molar mass of an element. Hence, smaller size molecule will have smaller mass. As a result, rate of diffusion will be more.
High temperature means more kinetic energy of molecules due to which more number of collisions will be there. Hence, rate of diffusion will also increase.
Thus, we can conclude that out of the given options a small concentration gradient is least likely to increase the rate of diffusion.
For i: 33mL
For ii: 87-88mL
For iii:22.3mL
Answer:
Boiling T° of solution = 100.6
Explanation:
Formula for elevation of boiling point is:
ΔT = Kb . m . i
where ΔT means Boiling T° of solution - Boiling T° of pure solvent
Our solute is a non ionizing compound.
i = 1, because it is a non ionizing compound. i, indicates the ions dissolved in solution.
m = molality (moles of solute dissolved in 1 kg of solvent)
90 g of solvent = 0.09 kg of solvent
We convert mass of solute to moles (by the molar mass):
10 g . 1 mol /92.09 g = 0.108 moles
m = 0.108 mol /0.09 kg = 1.21 m
Let's replace data: Boiling T° of solution - 100°C = 0.51 °C/m . 1.21 m . 1
Boiling T° of solution = 0.51 °C/m . 1.21 m . 1 + 100°C
Boiling T° of solution = 100.6
Find the mass of C in the 2.657 g CO2:
(2.657 g CO2) / (44.01 g/mol) = 0.06037 mol CO2
Since each mole of CO2 also has 1 mole of C, this is equivalent to 0.06037 mol C.
Find the mass of H in the 1.089 g H2O:
(1.089 g H2O) / (18.02 g/mol) = 0.06043 mol H2O
Since 1 mol H2O has 2 mol H, this is equivalent to (0.06043)*2 = 0.1209 mol H.
Taking the ratio of H to C: 0.1209 / 0.06037 = 2.002 ~ 2
Therefore, the empirical formula of isobutylene is CH2.
