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)
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8.8 × 10-5 M is the [H3O+] concentration in 0.265 M HClO solution.
Explanation:
HClO is a weak acid and does not completely dissociate in water as ions.
the equation of dissociation can be written and ice table to be formed.
HClO +H2O ⇒ ClO- + H3O+
I 0.265 0 0
C -x +x +x
E 0.265-x +x +x
Now applying the equation of Ka, where Ka is given as 2.9 × 10-8.
Ka =
2.9 × 10^-8 =
= 7.698 x
x = 8.8 × 10-5 M
The hydronium ion concentration is 8.8 × 10-5 M in 0.265 M solution of HClO.