Answer :
(1) The mass of silver nitrate is, 555 g
(2) The solubility of the gas will be, 0.433 g/L
Solution for Part 1 :
From the given data we conclude that
In 100 gram of water, the amount of silver nitrate = 222 g
In 250 gram of water, the amount of silver nitrate = [tex]\frac{222}{100}\times 250=555g[/tex]
Therefore, the mass of silver nitrate is, 555 g
Solution for Part 2 :
Formula used : [tex]S_1P_1=S_2P_2[/tex] (at constant temperature)
where,
[tex]S_1[/tex] = initial solubility of methane gas = 0.026 g/L
[tex]S_2[/tex] = final solubility of methane gas
[tex]P_1[/tex] = initial pressure of methane gas = 1 atm
[tex]P_2[/tex] = final pressure of methane gas = 0.06 atm
Now put all the given values in the above formula, we get the solubility of methane gas.
[tex](0.026g/L)\times (1atm)=S_2\times (0.06atm)[/tex]
[tex]S_2=0.433g/L[/tex]
Therefore, the solubility of the gas will be, 0.433 g/L
