Question

the molal freezing point depression constant for benzene is 4.9 0 Kelvin kg per mole Selenium exist as polymer of Selenium x, when 3.26 gram of Selenium is dissolved in 226 gram of benzene. the observed freezing point is 0.112 degree Celsius lower than for pure benzene decide the molecular formula of Selenium atomic weight of selenium is equal to 78. 8 gram​

Answer 1

Answer:

Δ$T_{f}$=$\frac{K_{f}w_{B}1000  }{w_{A}M_{B}  }$

Here

$K_{f}$=Molal depression constant

$w_{B}$=weight of the solute

$w_{A}$=weight of the solvent

$M_{B}$=Molar mass of the solute

substitute all the required values we get

273.112=$\frac{4.90*3.26*1000}{226*M_{B} }$

$M_{B}$=$\frac{4.90*3.26*1000}{226*0.112}$

$Se_{x} =631g$

79g*x=631g

=8

So the molecular formula for selenium is=$S_{8}$