Answer:
7 gram of Na2SO4 should be required to prepare 0.5L of a 0.100 M solution.
Explanation:
First of all the molecular weight of Na2SO4 is 142.08 gram.Now we all know that if the molecular weight of a compound is dissolved in 1000ml or 1 litee of water then the strength of that solution becomes 1 M.
According to the given question we have to prepare 0.100 M solution
1000 ml of solution contain 142.08×0.1= 14.208 gram Na2SO4
1 ml of solution contain 14.208÷1000= 0.014 gram
0.5L or 500ml of solution contain 0.014×500= 7gram Na2SO4.
So it can be stated that 7 gram of Na2SO4 should be required to prepare 0.5L of a 0.100M solution.
<u>Answer:</u> The value of <em>i</em> is 1.4 and 40% dissociation of 100 particles of zinc sulfate will yield 60 undissociated particles.
<u>Explanation:</u>
The equation used to calculate the Vant' Hoff factor in dissociation follows:

where,
= degree of dissociation = 40% = 0.40
i = Vant' Hoff factor
n = number of ions dissociated = 2
Putting values in above equation, we get:

The equation used to calculate the degee of dissociation follows:

Total number of particles taken = 100
Degree of dissociation = 40% = 0.40
Putting values in above equation, we get:

This means that 40 particles are dissociated and 60 particles remain undissociated in the solution.
Hence, 40% dissociation of 100 particles of zinc sulfate will yield 60 undissociated particles.
This process is called filtration. This process is a solid-fluid separation by the use of a medium wherein only the fluid (gases or liquids) can go through it. The medium is called the filter while the fluid that passed through the filter is called the filtrate. The solid particles are the large particles which cannot pass through the filter.
The answer to this is 22, confirmed by gradpoint