The freezing point of benzene c6h6 is 5.50°c at 1 atmosphere. a nonvolatile, nonelectrolyte that dissolves in benzene is tnt (trinitrotoluene). how many grams of tnt, c7h5n3o6 (227.1 g/mol), must be dissolved in 297.0 grams of benzene to reduce the freezing point by 0.500°c ?

The amount required in grams to reduce the freezing point by 0.500 degree Celsius will be 6.58 g.

Solution: equation to calculate depression in freezing point

given: = 0.500°C

=  for benzene is 5.12

m=?

y= TNT amount (in grams) required

now according to formula:

i think so it’s will be b yes i believe its