JEE MAIN - Chemistry (2002 - No. 20)
Freezing point of an aqueous solution is (-0.186)oC. Elevation of boiling point of the same solution is Kb = 0.512 oC, Kf = 1.86 oC, find the increase in boiling point.
0.186 oC
0.0512 oC
0.092 oC
0.2732 oC
Explanation
$$\Delta {T_b} = {K_b}{{{W_B}} \over {{M_B} \times {W_A}}} \times 1000;$$
$$\Delta {T_f} = {K_f}{{{W_B}} \over {{M_B} \times {W_A}}} \times 1000;$$
$${{\Delta {T_b}} \over {\Delta {T_f}}} = {{{K_b}} \over {{K_f}}} = {{\Delta {T_b}} \over { - 0.186}}$$
$$ = {{0.512} \over {1.86}}$$
$$ = {0.0512^ \circ }C.$$
$$\Delta {T_f} = {K_f}{{{W_B}} \over {{M_B} \times {W_A}}} \times 1000;$$
$${{\Delta {T_b}} \over {\Delta {T_f}}} = {{{K_b}} \over {{K_f}}} = {{\Delta {T_b}} \over { - 0.186}}$$
$$ = {{0.512} \over {1.86}}$$
$$ = {0.0512^ \circ }C.$$
Comments (0)
