Question

For the equilibrium $SrC{l}_2$. 6$H_{2}O$(s) $\leftrightharpoons$ $SrC{l}_2$.$2H_{2}O$(s) + 4$H_{2}O$(g)

the equilibrium constant $K_p$ = 16 $\times$ ${10}^{-12}$atm⁴ at $1^{\circ }$C. If one litre of air saturated with water vapour at $1^{\circ }$C is exposed to a large quantity of $SrC{l}_2$$2H_{2}O$(s), what weight of water vapour will be absorbed ? Saturated vapour pressure of water at $1^{\circ }$C = 7.6 torr.

• A. 6 g
• B. 2.3 g
• C. 6.4 g
• D. 8.5 g

Answer 1

The correct option is C

6.4 g

$SrC{l}_2$. 6$H_{2}O$(s) $\leftrightharpoons$ $SrC{l}_2$.$2H_{2}O$(s) + 4$H_{2}O$(g)

$K_p$ = ${\left(P_{H_{2}O}\right)}^4$ ; 16 $\times$ ${10}^{-12}$ = ${\left(P_{H_{2}O}\right)}^4$ ; $P_{H_{2}O}$ = 0.002 atm = 1.52 torr.

Number of moles of $H_{2}O$ vapour in 1 liter of air at equilibrium

$n_{H_{2}O}$ = $\frac{PV}{RT}$ = $\frac{0.002\times 1}{0.0821\times 274}$ = 8.89 $\times$ ${10}^{-5}$

Saturated vapour pressure $P_{H_{2}O}$ = 7.6 torr = $\frac{7.6}{760}$ = 0.01 atm.

Number of moles of $H_{2}O$ vapour in 1 liter of saturated air.

$n_{H_{2}O}$ = $\frac{PV}{RT}$ = $\frac{0.01\times 1}{0.0821\times 274}$ = 4.44 $\times$ ${10}^{-4}$

Mass of $H_{2}O$ vapour absorbed = (3.554 $\times$ ${10}^{-4}$) $\times$ 18 = 0.00639g = 6.39g