CBSE Class 12 · 2003 · All India · SolutionsShort Answer

An aqueous solution containing 1.248 g of barium chloride (molar mass = 208.34 g mol

^{- 1}

) in 100 g of water boils at 100.0832°C. Calculate the degree of dissociation of barium chloride. (K

_{b}

of water = 0.52 K kg mol

^{- 1}

)

Answer: α ≈ 0.835 (83.5%)

Solution

Molality:

m = \frac{1.248/208.34}{0.100} = 0.0599 mol kg^{- 1}

Calculated elevation (no dissociation):

Δ T_{b}^{c a l c} = 0.52 \times 0.0599 = 0.03115 K

Observed elevation = 0.0832 K, so

i = \frac{0.0832}{0.03115} = 2.67

BaCl

_{2} \to

^{2 +}

+ 2Cl

^{-}

gives

n = 3

ions:

i = 1 + 2 α \Rightarrow α = \frac{2.67 - 1}{2}

α = 0.835 (83.5% dissociated)

CBSE Class 12 Board · 2003 · All India · Subjective