Asolution is prepared by condensing 4.00 l of a gas, measured at 27°c and 748 mmhg pressure, into 58.0 g of benzene. calculate the freezing point of this solution. [kfp(benzene) = 5.12°c/m, kbp(benzene) = 2.53°c/m] (the boiling point and freezing point of benzene are 80.1°c and 5.5°c, respectively).

First, we are using the ideal gas law to get n the number of moles:

PV = nRT

when P is the pressure = 748 mmHg/760 = 0.984 atm

V is the volume = 4 L

R is ideal gas constant = 0.0821

T is the temperature in Kelvin = 300 K

∴ n =  0.984atm*4L/0.0821*300

       = 0.1598 moles

when the concentration = moles * (1000g / mass)

                                         = 0.1598 * (1000g / 58 g )

                                         = 2.755 M

when the freezing point = 5.5 °C

and Kf = - 5.12 °C/m

∴ the freezing point for the solution = 5.5 °C + (Kf*m)

                                                            = 5.5 °C - (5.12°C/m * 2.755m)

                                                            = -8.6 °C

What record? the question is incomplete

The se-ci  bonds is more polar  than s-br bond