C6H6 freezes at 5.5°C. The temperature at which a solution of 10 g o

1.	C6H6 freezes at 5.5°C. The temperature at which a solution of 10 g of C4H10 in 200 g of C6H6 freeze is ...........°C.
Answer» We have given, Freezing point of pure C₆H₆ = 5.5\(^\circ\) C The molal freezing point depression constant = 5.12\(^\circ\) C/m = k_f \(\triangle\)T_f= K_fm Where, \(\triangle\)T_f= depression in freezing point m = molality of solution K_f= Depression constant. Number of moles of C₄H₁₀ = \(\cfrac{4g}{58}\) = 0.069 mol molality of solution = \(\cfrac{0.069}{200}\times1000\) = 0.345 m \(\therefore\) \(\triangle\)T_f= 5.12 x 0.345 = 1.77\(^\circ\) C \(\therefore\) Freezing point of solution will be = 5.5\(^\circ\) C - 1.8\(^\circ\) C = 3.7\(^\circ\) C

C6H6 freezes at 5.5°C. The temperature at which a solution of 10 g of C4H10 in 200 g of C6H6 freeze is ...........°C.

Answer»

We have given,

Freezing point of pure C₆H₆ = 5.5\(^\circ\) C

The molal freezing point depression constant = 5.12\(^\circ\) C/m = k_f

\(\triangle\)T_f= K_fm

Where, \(\triangle\)T_f= depression in freezing point

m = molality of solution

K_f= Depression constant.

Number of moles of C₄H₁₀ = \(\cfrac{4g}{58}\)

= 0.069 mol

molality of solution = \(\cfrac{0.069}{200}\times1000\)

= 0.345 m

\(\therefore\) \(\triangle\)T_f= 5.12 x 0.345

= 1.77\(^\circ\) C

\(\therefore\) Freezing point of solution will be = 5.5\(^\circ\) C - 1.8\(^\circ\) C

= 3.7\(^\circ\) C

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