JEE MAIN - Chemistry (2003 - No. 30)
In a 0.2 molal aqueous solution of a weak acid HX the degree of ionization is 0.3. Taking kf for water as 1.85, the freezing point of the solution will be nearest to
-0.360oC
-0.260oC
+0.480oC
-0.480oC
Explanation
$$\Delta {T_f} = {K_f} \times m \times i;$$
$$\Delta {T_f} = 1.855 \times 0.2 \times 1.3 = {0.480^ \circ }C$$
$$\therefore$$ $$\,\,\,{T_f} = 0 - {0.480^ \circ }C = - {0.480^ \circ }C$$
$$\mathop {\left( {HX} \right.}\limits_{1 - 0.3} \,\,\rightleftharpoons\,\,\mathop {{H^ + }}\limits_{0.3} + \mathop {{X^ - }}\limits_{0.3} ,i = 1.\left. 3 \right)$$
$$\Delta {T_f} = 1.855 \times 0.2 \times 1.3 = {0.480^ \circ }C$$
$$\therefore$$ $$\,\,\,{T_f} = 0 - {0.480^ \circ }C = - {0.480^ \circ }C$$
$$\mathop {\left( {HX} \right.}\limits_{1 - 0.3} \,\,\rightleftharpoons\,\,\mathop {{H^ + }}\limits_{0.3} + \mathop {{X^ - }}\limits_{0.3} ,i = 1.\left. 3 \right)$$
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