henrys-law

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Q1.

Which of the following graph correctly represents the plots of \(\mathrm{K}_{\mathrm{H}}\) at 1 bar for gases in water versus temperature?

As temperature increases solubility first decrease then increase hence \(\mathrm{K}_{\mathrm{H}}\) first increase than decrease also at moderate temperature \(\mathrm{K}_{\mathrm{H}}\) value \(\mathrm{He}>\mathrm{N}_2>\) \(\mathrm{CH}_4\).

Q2.

Sea water, which can be considered as a 6 molar \((6 \mathrm{M})\) solution of NaCl , has a density of \(2 \mathrm{~g} \mathrm{~mL}^{-1}\). The concentration of dissolved oxygen \(\left(\mathrm{O}_2\right)\) in sea water is 5.8 ppm . Then the concentration of dissolved oxygen \(\left(\mathrm{O}_2\right)\) in sea water, is \(x \times 10^{-4} \mathrm{~m}\).

\(x=\) ___________. (Nearest integer)

Given: Molar mass of NaCl is \(58.5 \mathrm{~g} \mathrm{~mol}^{-1}\)

Molar mass of \(\mathrm{O}_2\) is \(32 \mathrm{~g} \mathrm{~mol}^{-1}\)

Sea water is 6 Molar in NaCl , So 1000 ml of sea water contains 6 mol of NaCl .

\(\begin{aligned} & \text { mass of solution }=\text { Volume } \times \text { density } \\ & =1000 \times 2 \\ & \text { mass of solution }=2000 \mathrm{~g} \\ & \mathrm{ppm}=\frac{\text { mass of } \mathrm{O}_2}{2000} \times 10^6 \\ & \text { mass of } \mathrm{O}_2=5.8 \times 2 \times 10^{-3} \\ & \quad=1.16 \times 10^{-2} \mathrm{~g} \\ & \text { molality for } \mathrm{O}_2=\frac{1.16 \times 10^{-2} / 32}{(2000-6 \times 58.5)} \times 1000 \end{aligned}\)

\(\begin{aligned} &\begin{aligned} & =\frac{1.16 \times 10}{32 \times 1649} \\ & =0.000219 \\ & =2.19 \times 10^{-4} \end{aligned}\\ &\text { Correct answer } \Rightarrow 2 \end{aligned}\)

Q3.

A company dissolves 'x' amount of CO₂ at 298 K in 1 litre of water to prepare soda water. X = __________ \(\times\) 10^\(-\)3 g. (nearest integer)

(Given : partial pressure of CO₂ at 298 K = 0.835 bar.

Henry's law constant for CO₂ at 298 K = 1.67 kbar.

Atomic mass of H, C and O is 1, 12, and 6 g mol^\(-\)1, respectively)

According to Henry's law, partial pressure of a gas is given by

\(P_{g}=\left(K_{H}\right) X_{g}\)

where \(X_{g}\) is mole fraction of gas in solution

\(0.835=1.67 \times 10^{3}\left(\mathrm{X}_{\mathrm{CO}_{2}}\right)\)

\(\mathrm{X}_{\mathrm{CO}_{2}}=5 \times 10^{-4}\)

Mass of \(\mathrm{CO}_{2}\) in \(1 \mathrm{~L}\) water \(=1221 \times 10^{-3} \mathrm{~g}\)

Q4.

For a solution of the gases A, B, C and D in water at 298 K, the values of Henry's law constant (K_H) are 30.40, 2.34, 1.56 \(\times\) 10^\(-\)5 and 0.513 k bar respectively. In the given graph, the lines marked as 'p' and 's' correspond respectively to :

JEE Main 2022 (Online) 30th June Morning Shift Chemistry - Solutions Question 71 English

A and C
B and A
D and A
C and D

According to Henry's law \(p={K_{H}} \times x\)

for \(x\) constant \(: p ~ \alpha ~ K_{\mathrm{i}}\)

\(p\) has max partial pressures.

So it should have maximum \(K_{\mathrm{H}}\) and \(s\) have minimum partial pressure, So \(K_{H}\) should be minimum.

Q5.

If \(\mathrm{O}_{2}\) gas is bubbled through water at \(303 \mathrm{~K}\), the number of millimoles of \(\mathrm{O}_{2}\) gas that dissolve in 1 litre of water is __________. (Nearest Integer)

(Given : Henry's Law constant for \(\mathrm{O}_{2}\) at \(303 \mathrm{~K}\) is \(46.82 \,\mathrm{k}\) bar and partial pressure of \(\mathrm{O}_{2}=0.920\) bar)

(Assume solubility of \(\mathrm{O}_{2}\) in water is too small, nearly negligible)

\(\mathrm{P}=\mathrm{K}_{\mathrm{H}} \times \mathrm{X}\)

\(0.920 \mathrm{bar}=46.82 \times 10^{3}\, \mathrm{bar} \times \frac{\mathrm{mol} \,\mathrm{of} \,\mathrm{O}_{2}}{\mathrm{~mol} \text { of } \mathrm{H}_{2} \mathrm{O}}\)

\(0.920=46.82 \times 10^{3} \times \frac{\text { mol of O}_{2}}{1000 / 18}\)

\(0.920=46.82 \times n_{O_{2}}\)

\(\mathrm{P}=\frac{0.920}{46.82 \times 18}=n_{O_{2}}\)

\(\Rightarrow 1.09 \times 10^{-3} \,n_{O_{2}}\)

\(\Rightarrow \mathrm{m}\, \mathrm{mol}\, \mathrm{of}\, \mathrm{O}_{2}=1\)

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