Question

at 298 K and 1 atm pressure , solubility of N2 in water was found to be 6.8*10-4 mol/L . if the partial pressure of N2 is 0.78 atm . then what is the concentration of N2 dissolved in water under atmospheric conditions

Answer 1

Answer:

5.3 ×10^-4

Explanation:

Initial Partial Pressure of N2, p=1 atm

Using p = K × Concentration

1 = K × 6.8 × 10^-4

K = 1.47 10^3

Final partial pressure of N2, P = 0.78

Again,

0.78 = 1.47 × 10^3 × Concentration(c)

Hence, c= 5.3 × 10^-4

Answer 2

The concentration of N2 dissolved in water under atmospheric pressure is $5.3\times 10-^{4}mol/l$.

Given:

$298k$ and $1 atm$.

Solubility in water = $6.8\times 10-^{4}$.

The pressure of N2 = $0.78atm$.

To Find:

The concentration of N2 in water.

Solution:

Here, it is given that the solubility in water is $6.8\times 10-^{4}$.

now, we have to apply Henry's law, which is $p=KHC$.

here,

$p=$ partial pressure.

$KH=$ Henry's constant.

and,

$C=$ Concentration.

So, here it is given that the solubility of N, at 1 atm $6.8\times 10-^{4}$$mol/L$.

here, $p=1atm$

so, $KH$ will be equal to $p/C$ $=\frac{1}{6.8}\times 10-^{4}$.

after calculating the above equation.

we get,

$KH=1.47\times 10^{3}atm/mol$.

so, the partial pressure $p=0.78atm$.

now,

$C=\frac{P}{K_h}$

here, after putting the given values in the above law.

we get,

$C=\frac{0.78}{1.47\times 10^{3} }\\ =0.53\times 10-^{3}\\ =5.3\times 10-^{4}mol/l$.

Hence, the concentration of N2 dissolved in water under atmospheric pressure is $5.3\times 10-^{4}mol/l$.

                                                                                          #SPJ3