The vapour pressure of a solvent at 293 K is 100 mm Hg. Then the vapour pressure of a solution containing 1 mole of a strong electrolyte (AB2) in 99 moles of the solvent at 293 K is (assume complete dissociation of solute)

Question

The vapour pressure of a solvent at 293 K is 100 mm Hg. Then the vapour pressure of a solution containing 1 mole of a strong electrolyte (AB2) in 99 moles of the solvent at 293 K is (assume complete dissociation of solute) (A) 103 mm Hg (B) 99 mm Hg (C) 97 mm Hg (D) 101 mm Hg

Tardigrade Admin · Accepted Answer

Let us consider the problem: (( P A 0- P 0)/ P A 0)= i (( n B )/( n A + n B )) Since, ((100-PA)/100)=3[(1/(99+1))] Since, 100- P A =3 Hence the solution is P A =97 mm of Hg.

Q. The vapour pressure of a solvent at 293 K is 100 mm Hg. Then the vapour pressure of a solution containing 1 mole of a strong electrolyte $(A B_{2})$ in 99 moles of the solvent at 293 K is (assume complete dissociation of solute)

103 mm Hg

99 mm Hg

97 mm Hg

101 mm Hg

98 mm Hg

Let us consider the problem:
$\frac{( P _{A} ^{0} - P ^{0} )}{P _{A} ^{0}} = i \frac{( n _{B} )}{( n _{A} + n _{B} )}$
Since,
$\frac{( 100 - P _{A} )}{100} = 3 [\frac{1}{( 99 + 1 )}]$
Since,
$100 - P_{A} = 3$
Hence the solution is $P_{A} = 97 mm$ of $H g$ .

Q. The vapour pressure of a solvent at 293 K is 100 mm Hg. Then the vapour pressure of a solution containing 1 mole of a strong electrolyte (AB2​) in 99 moles of the solvent at 293 K is (assume complete dissociation of solute)