CBSE Class 12 Chemistry 2016 Outside Delhi Set 1 Solved Paper

Question : 15 of 26

Marks: +1, -0

Calculate the boiling point of solution when

4 {\g}

{\MgSO}_4 ( {\M}=120 {\g} {\mol}^{-1})

was dissolved in 100

{\g}

of water, assuming

{\MgSO}_4

undergoes complete ionization.

( {\K}_b

for water

=0.52 {\K} {\kg} {\mol}^{-1})

Solution:

∆ {\T}_b\;=i {\K}_b ⋅ {\m}

i\;=2

\;=i × {\K}_b × \;{ {\W}_2 × 1000}/{ {\M} × {\W}_1}

\;=2 × 0.52 {\K} {\kg} {\mol}^{-1}

× \;{4 {\g} × 1000 {\g} \/ {\kg}}/{120 {\g} \/ {\mol} × 100 {\g}}

\;=\;{2 × 0.52}/{3}

\;=0.346 {\K}

\text " Boiling point of water "\;=\;{373.15 {\K}}\/{373 {\K}}

{\T}_b\;= {\T}_b^0+∆ {\T}_b

\;=373.15 {\K}+0.346 {\K} \;\text " or "\;

373 {\K}+0.346 {\K}

\;=373.496 {\K} \;\text " or "\; 373.346 {\K}

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