Answer: Freezing point of a solution will be 
Explanation:
Depression in freezing point is given by:
= Depression in freezing point
i= vant hoff factor = 1 (for non electrolyte)
= freezing point constant = 
m= molality
Weight of solvent (benzene)= 1480 g =1.48 kg
Molar mass of solute (octane) = 114.0 g/mol
Mass of solute (octane) = 220 g
Thus the freezing point of a solution will be
Answer:
53.5g of NH4Cl
Explanation:
First, we need to obtain the number of mole of NH4Cl. This is illustrated below:
Volume = 0.5L
Molarity = 2M
Mole =?
Molarity = mole /Volume
Mole = Molarity x Volume
Mole = 2 x 0.5
Mole = 1mole
Now, let us convert 1mole of NH4Cl to gram. This is illustrated below:
Molar Mass of NH4Cl = 53.5g/mol
Number of mole = 1
Mass =?
Number of mole = Mass /Molar Mass
Mass = number of mole x molar Mass
Mass = 1 x 53.5
Mass = 53.5g
Therefore, 53.5g of NH4Cl is contained in the solution.
Answer:
Avogadro's number or Avogardro’s constant
Explanation:
I’m pretty sure this is correct if it’s not I’m sorry lol.
Answer:
option C is correct = 1.14 × 10²² molecules of CO₂
Explanation:
Given data:
Number of moles of CO₂ = 0.0189 mol
Number of molecules = ?
Solution:
The given problem will solve by using Avogadro number.
It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.
The number 6.022 × 10²³ is called Avogadro number.
For example,
18 g of water = 1 mole = 6.022 × 10²³ molecules of water
For given question:
1 mole of CO₂ = 6.022 × 10²³ molecules of CO₂
0.0189 mol of CO₂ × 6.022 × 10²³ molecules of CO₂ / 1mol
1.14 × 10²² molecules of CO₂
Thus, option C is correct.
