Chemistry | TS EAMCET 2018 | Discussion Page

Henry's law constant for $CO_{2}$ in water is 1.67 K bar at $25^{0}C$ . The quantity of $CO_{2}$ in 1000 mL of soda water when packed under 5 bar $CO_{2}$ pressure at $25^{0} C$ is

A) 0.084 mol

B) 0.167 mol

C) 0.252 mol

D) 0.336 mol

Answer:

Option B

Explanation:

We know that

$p= K_{B}\chi_{i}\Rightarrow p=K_{H}\times \frac{n_{1}}{n_{1}+n_{2}}$

Given is p= pressure (5 bar)

$K_{H}$ = Henery's constant (1.67 K bar)

$\chi_{1}$ = mole fraction of solute $CO_{2}$

$\chi_{1}$ = $\frac{n_{1}}{n_{1}+n_{2}} \Rightarrow =5=1.67 \times 10^{3} \times \frac{ H_{1}}{55.5}$

$H_{1}= \frac{55.5 \times 5}{1.67 \times 10^{3}}=0.167 mol$

Discussion Forum

Answer:

Explanation:

χ1\chi_{1}= n1n1+n2⇒=5=1.67×103×H155.5\frac{n_{1}}{n_{1}+n_{2}} \Rightarrow =5=1.67 \times 10^{3} \times \frac{ H_{1}}{55.5}

$\chi_{1}$ = $\frac{n_{1}}{n_{1}+n_{2}} \Rightarrow =5=1.67 \times 10^{3} \times \frac{ H_{1}}{55.5}$