A company dissolves ' X ' amount of CO 2 at 298 K in 1 litre of water to prepare soda water X = × 10-3 g . (nearest integer) (Given: partial pressure of CO 2 at 298 K =0.835 bar. Henry's law constant for CO 2 at 298 K =1.67 kbar. Atomic mass of H , C and O is 1,12 and 6 g mol -1, respectively)

Question

A company dissolves ' X ' amount of CO 2 at 298 K in 1 litre of water to prepare soda water X = × 10-3 g . (nearest integer) (Given: partial pressure of CO 2 at 298 K =0.835 bar. Henry's law constant for CO 2 at 298 K =1.67 kbar. Atomic mass of H , C and O is 1,12 and 6 g mol -1, respectively) (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

From Henry law P = K H X CO 2 0.835=1.67 × 103 × 1.67 × 103 × ( w CO 2 / 44/( w CO 2)44+(1000)18 w / CO 2)=1.2228 g =1222.8 × 10-3 g Or P = K H X CO 2 0.835=1.67 × 103 × ( n CO 2/ n CO 2+ n H 2 O ) 0.835=1.67 × 103 × ( w CO 2 / 44/(1000)18) w CO 2=1.2222 g =1222.2 × 10-3 g

Q. A company dissolves ' X ' amount of CO2​ at 298K in 1 litre of water to prepare soda water X= ______×10−3g. (nearest integer) (Given: partial pressure of CO2​ at 298K=0.835 bar. Henry's law constant for CO2​ at 298K=1.67kbar. Atomic mass of H,C and O is 1,12 and 6gmol−1, respectively)

From Henry law P=KH​XCO2​​ 0.835=1.67×103×1.67×103×44wCO2​​​+181000​wCO2​​/44​ wCO2​​=1.2228g=1222.8×10−3g Or P=KH​XCO2​​ 0.835=1.67×103×nCO2​​+nH2​O​nCO2​​​ 0.835=1.67×103×181000​wCO2​​/44​ wCO2​​=1.2222g=1222.2×10−3g