Read the statement.
1 answer:
the final concentration of NaI solution in 60 grams/litre.
Explanation:
Given that:
Initial concentration of NaI solution M1 = 0.2 M
initial volume of NaI V1 = 2 L
Final volume V2 = 1 Litre
Final molarity=?
concentration in grams/litre = ?
molar mass of NaI = 150 gram/mole
For dilution following formula is used:
M1 V1 = M2V2
putting the values in the equation
0.2 X 2 = 1 X M2
M2 = 0.4
For concentration in grams/litre, formula used
molarity =
mass = 0.4 x 150
= 60 grams
So, 60 grams of NaI will be present in final solution of NaI after evaporation.
The concentration is 60 grams/ L (as volume got reduced to 1 litre from 2 litres)
You might be interested in
Answer:
A) increasing dispersion interactions
Explanation:
Polarizability allows gases containing atoms or nonpolar molecules (for example, to condense. In these gases, the most important kind of interaction produces <em>dispersion forces</em>, <em>attractive forces that arise as a result of temporary dipoles induced in atoms or molecules.</em>
<em>Dispersion forces</em>, which are also called <em>London forces</em>, usually <u>increase with molar mass because molecules with larger molar mass tend to have more electrons</u>, and <u>dispersion forces increase in strength with the number of electrons</u>. Furthermore, larger molar mass often means a bigger atom whose electron distribution is more easily disturbed because the outer electrons are less tightly held by the nuclei.
Because the noble gases are all nonpolar molecules, <u>the only attractive intermolecular forces present are the dispersion forces</u>.
