An aqueous solution containing 1.248 g of barium chloride (molar mass =208.34g mol-1 ) in 100 g of water boils at 100.0832oC . The degree of dissociation of BaCl2 will be ( Kb for water =0.52K text kg text mol-1 )

Question

An aqueous solution containing 1.248 g of barium chloride (molar mass =208.34g mol-1 ) in 100 g of water boils at 100.0832oC . The degree of dissociation of BaCl2 will be ( Kb for water =0.52K text kg text mol-1 ) (A)  0.835  (B)  0.756  (C)  0.638  (D)  0.930

Tardigrade Admin · Accepted Answer

M2 (observe =(1000× 0.52× 1.248/100× 0.0832)=78g mol-1 i=(208.34/78)=2.67 underset1-α mathop underset1 mol mathopBaCl2 xrightarrow undersetα mathop underset0 mathopBa2+ + underset beginsmallmatrix 0 2α endsmallmatrix mathop2Cl- Number of total moles =1+2α or i=1+2α or α =(i-1/2)=(2.67-1/2)=0.835

Q. An aqueous solution containing 1.248 g of barium chloride (molar mass $= 208.34 g m o l^{- 1}$ ) in 100 g of water boils at $100.0832^{o} C$ . The degree of dissociation of $B a C l_{2}$ will be ( $K_{b}$ for water $= 0.52 K k g m o l^{- 1}$ )

$0.835$

$0.756$

$0.638$

$0.930$

$M_{2}$ (observe $= \frac{1000 \times 0.52 \times 1.248}{100 \times 0.0832} = 78 g m o l^{- 1}$ $i = \frac{208.34}{78} = 2.67$ $1 - α 1 m o l B a C l_{2} α 0 B a^{2 +} + 0 2 α 2 C l^{-}$ Number of total moles $= 1 + 2 α$ or $i = 1 + 2 α$ or $α = \frac{i - 1}{2} = \frac{2.67 - 1}{2} = 0.835$

Q. An aqueous solution containing 1.248 g of barium chloride (molar mass =208.34gmol−1 ) in 100 g of water boils at 100.0832oC . The degree of dissociation of BaCl2​ will be ( Kb​ for water =0.52Kkgmol−1 )

0.835

0.756

0.638

0.930