Henry's law constant for N₂ at 293 K is 76.48 kbar. If N₂ is bubbled through water at 293 K at a pressure of 1 atm . Then how many milimoles of it will dissolve in 10 ltr of water.

Given - 
Temperature = 293 K
Henry's law constant  = 76.48 kbar
​pressure = 1 atm
Calculation -
The solubility of any gas is related to the mole fraction which can be calculated by applying Henry's law.
x =$\frac{p}{K_H}$
where, x = mole fraction of nitrogen gas
p = pressure of the gas
K_H​ = henry's constant
now, pressure is given in atm and Henry's constant in bar. We have to convert atm in bar
1 atm = 1.013 bar 
x = $\frac{1.013 bar}{76.48 \times {10}^3 bar}$
x = 0.01325 $\times$ 10^-3
x = 1.325 $\times$ 10^-5
mole fraction, x of nitrogen  = $\frac{n \left(nitrogen\right)}{n\left(nitrogen\right) + n\left(water\right)}$
where n = number of moles. 
number of moles of water in 10 L of water  = mass of 10 L of water/ molar mass
Assuming density of water  = 1 gcm​^-3
Density  = mass / volume
1 L = 1000 cm^3
​So, 10 L = 10000 cm³
1 gcm^​-3 = mass / 10000 cm³
Thus, mass  = 10000 g.
Number of moles of water  = 10000 g / 18 gmol^-1
number of moles = 555.56 moles
So, mole fraction of nitrogen  = $\frac{n\left(nitrogen\right)}{n\left(nitrogen\right) + n\left(water\right)}$
 1.325 $\times$ 10^-5   = $\frac{n}{n + 555.56}$
neglecting n (nitrogen) in denominator with respect to 555.56
we get, 
 1.325 $\times$ 10^-5  = $\frac{n}{555.56}$
n = 1.325 $\times$ 10^-5 $\times$ 555.6
n = 736.117 $\times$ 10^-5
n = 7.36 $\times$ 10^-3
n = 7.36 milimoles.