CO 2 gas is bubbled through water during a soft drink manufacturing process at 298 K. If CO 2 exerts a partial pressure of 0.835 bar then x m mol of CO 2 would dissolve in 0.9 L of water. The value of x is . (Nearest integer) (Henry's law constant for CO 2 at 298 K is 1.67 × 103 bar )

Question

CO 2 gas is bubbled through water during a soft drink manufacturing process at 298 K. If CO 2 exerts a partial pressure of 0.835 bar then x m mol of CO 2 would dissolve in 0.9 L of water. The value of x is . (Nearest integer) (Henry's law constant for CO 2 at 298 K is 1.67 × 103 bar ) (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

From Henry's law P text gas =KH ⋅ X text gas 0.835=1.67 × 103 × (n(C2)/(0.9 × 1000)18) n( CO 2)=0.025 Millimoles of CO 2=0.025 × 1000=25

Q. CO2​ gas is bubbled through water during a soft drink manufacturing process at 298K. If CO2​ exerts a partial pressure of 0.835 bar then xmmol of CO2​ would dissolve in 0.9L of water. The value of x is . (Nearest integer) (Henry's law constant for CO2​ at 298K is 1.67×103 bar )

From Henry's law Pgas ​=KH​⋅Xgas ​ 0.835=1.67×103×180.9×1000​n(C2​)​ n(CO2​)=0.025 Millimoles of CO2​=0.025×1000=25

From Henry's law
$P_{gas} = K_{H} \cdot X_{gas}$
$0.835 = 1.67 \times 1 0^{3} \times \frac{n ( C _{2} )}{\frac{0.9 \times 1000}{18}}$
$n (C O_{2}) = 0.025$
Millimoles of $C O_{2} = 0.025 \times 1000 = 25$