JEE MAIN - Chemistry (2021 - 24th February Evening Shift - No. 19)
C6H6 freezes at 5.5$$^\circ$$C. The temperature at which a solution of 10g of C4H10 in 200g of C6H6 freeze is __________ $$^\circ$$C. (The molal freezing point depression constant of C6H6 is 5.12$$^\circ$$C/m.)
Answer
1
Explanation
Pure solvent : C6H6(l)
Given,
$$T_f^o = 5.5^\circ C$$
$${K_f} = 5.12^\circ C/m \Rightarrow m = 200g$$
$${m_{solute}} = 10g$$
Molar mass of solute $${C_4}{H_{10}} = 12 \times 4 + 10 = 58$$
Solute (C4H10) is non-dissociative;
$$\therefore$$ i = 1
$$\therefore$$ $$\Delta {T_f} = i{K_f}\,m$$
$$ \Rightarrow (T_f^o - T_f^1) = 1 \times 5.12 \times {{(10/58)} \over {(200/1000)}}$$
$$5.5 - T_f^1 = {{5.12 \times 5 \times 10} \over {58}} \Rightarrow T_f^1 = 1.086^\circ C$$
or $$T_f^1 \approx 1^\circ C$$
Given,
$$T_f^o = 5.5^\circ C$$
$${K_f} = 5.12^\circ C/m \Rightarrow m = 200g$$
$${m_{solute}} = 10g$$
Molar mass of solute $${C_4}{H_{10}} = 12 \times 4 + 10 = 58$$
Solute (C4H10) is non-dissociative;
$$\therefore$$ i = 1
$$\therefore$$ $$\Delta {T_f} = i{K_f}\,m$$
$$ \Rightarrow (T_f^o - T_f^1) = 1 \times 5.12 \times {{(10/58)} \over {(200/1000)}}$$
$$5.5 - T_f^1 = {{5.12 \times 5 \times 10} \over {58}} \Rightarrow T_f^1 = 1.086^\circ C$$
or $$T_f^1 \approx 1^\circ C$$
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