One mole of an ideal gas at 300 K and 20 atm expands to 2 atm under isothermal and reversible conditions. The work done by the gas is $-x \mathrm{~kJ} \mathrm{~mol}^{-1}$. The value of $x$ is $\left(R=8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right)$
At 1000 K , the equilibrium constant for the reaction, $\mathrm{CO}_2(\mathrm{~g})+\mathrm{H}_2(\mathrm{~g}) \rightleftharpoons \mathrm{CO}(\mathrm{g})+\mathrm{H}_2 \mathrm{O}(\mathrm{g})$ is 0.53 . In a one litre vessel, at equilibrium the mixture contains 0.25 mole of $\mathrm{CO}, 0.5$ mole of $\mathrm{CO}_2, 0.6$ mole of $\mathrm{H}_2$ and $x$ moles of $\mathrm{H}_2 \mathrm{O}$. The value of $x$ is
$$ \text { Match the following. } $$
| $$ \text { List I (Reactions) } $$ |
$$ \text { List II (Methods) } $$ |
||
|---|---|---|---|
| (A) | $$ \begin{aligned} \mathrm{Mg}\left(\mathrm{HCO}_3\right)_2 \longrightarrow \mathrm{Mg}(\mathrm{OH})_2 \downarrow & +2 \mathrm{CO}_2 \uparrow \end{aligned} $$ |
(I) | Clark's method |
| (B) | $$ \begin{array}{r} M^{2+}+\mathrm{Na}_4 \mathrm{P}_6 \mathrm{O}_{18}^{2-} \longrightarrow\left[\mathrm{Na}_2 \mathrm{MP}_6 \mathrm{O}_{18}\right]^{2-} +2 \mathrm{Na}^{+} \end{array} $$ |
(II) | Ion exchange method |
| (C) | $$ \begin{aligned} \mathrm{Ca}\left(\mathrm{HCO}_3\right)_2 & +\mathrm{Ca}(\mathrm{OH})_2 \longrightarrow 2 \mathrm{CaCO}_3+2 \mathrm{H}_2 \mathrm{O} \end{aligned} $$ |
(III) | Boiling |
| (D) | $$ \begin{gathered} 2 \mathrm{NaZ}+\mathrm{Ca}^{2+}(\mathrm{aq}) \longrightarrow 2 \mathrm{Na}^{+}+\mathrm{CaZ} (\mathrm{Z}=\text { Zeolite }) \end{gathered} $$ |
(IV) | Calgon's method |
The correct answer is
Observe the following statements.
Statement I Both LiF and CsI have low solubility in water.
Statement II Low solubility of LiF in water is due to smaller hydration enthalpy of ions and that of CsI is due to its high lattice enthalpy.
The correct answer is
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