For the solution of the gases w,x,y and z in | vedclass

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(C) According to Henry's law,the partial pressure of a gas $(P_{gas})$ is related to its mole fraction $(X_{gas})$ as: $P_{gas} = K_H \cdot X_{gas}$.S

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(C) According to Henry's law,the partial pressure of a gas $(P_{gas})$ is related to its mole fraction $(X_{gas})$ as: $P_{gas} = K_H \cdot X_{gas}$.Since $X_{gas} + X_{H_2O} = 1$,we have $X_{gas} = 1 - X_{H_2O}$.Substituting this into the equation: $P_{gas} = K_H(1 - X_{H_2O}) = K_H - K_H \cdot X_{H_2O}$.This is a linear equation of the form $y = mx + c$,where $y = P_{gas}$,$x = X_{H_2O}$,slope $m = -K_H$,and intercept $c = K_H$.As the mole fraction of water $(X_{H_2O})$ increases,the partial pressure of the gas decreases linearly.The intercept on the y-axis is $K_H$. The values of $K_H$ are $0.5$,$2$,$35$,and $40 \ kbar$ for $w$,$x$,$y$,and $z$ respectively.Therefore,the intercept increases in the order $w The slope is $-K_H$,meaning the lines have negative slopes,and the magnitude of the slope increases as $K_H$ increases.Looking at the options,the graph in Option $C$ shows lines with negative slopes where the y-intercepts increase in the order $w < x < y < z$.

For the solution of the gases $w$,$x$,$y$ and $z$ in water at $298 \ K$,the Henry's law constants $(K_H)$ are $0.5$,$2$,$35$ and $40 \ kbar$,respectively. The correct plot for the given data is

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