CBSE Class 12 Chemistry 2018 Solved Paper

Question : 6 of 26

Marks: +1, -0

Calculate the freezing point of a solution containing

60 {\g}

of glucose (Molar mass

=180 {\g} {\mol}^{-1}

) in

250 {\g}

of water.

( {\K}_{ {\f}}

of water

=1.86 {\K} {\kg} {\mol}^{-1}) \;\; 2

Solution:

∆ T_f\;=K_f m

\;=K_f × \;{w_2 × 1000}/{M_2 × w_1}

\;=\;{1.86 × 60 × 1000}/{180 × 250}

\;=2.48 {\K}

∆ T_f\;=T_f^o-T_f

2.48\;=273.15-T_f

{\T}_f\;=273.15-2.48=270.67 {\K}

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