Determine the freezing point depression of a solution that contains 30.7 g glycerin (c3h8o3, molar mass = 92.09 g/mol) in 376 ml of water. some possibly useful constants for water are kf = 1.86°c/m and kb = 0.512°c/m.

Answer : The freezing depression of a solution is

Explanation :  Given,

Molal-freezing-point-depression constant for water =

Mass of glycerin (solute) = 30.7 g

Volume of water (solvent) = 376 mL

Molar mass of glycerin = 92.09 g/mole

First we have to calculate the mass of water.

As we know that, the density of water is 1.0 g/mL.

So, mass of water = Density of water × Volume of water

Mass of water = 1.0 g/mL × 376 mL

Mass of water = 376 g = 0.376 kg   (1 g = 0.001 kg)

Now we have to calculate the freezing point depression of a solution.

Formula used :  

where,

= change in freezing point

= freezing point of solution = ?

= freezing point of water =

i = Van't Hoff factor = 1  (for glycerin non-electrolyte)

= freezing point constant for water =

m = molality

Now put all the given values in this formula, we get

Therefore, the freezing depression of a solution is

answer: e.   sour taste

explanation: acids are those substances which dissociates in water to give ions. as they dissociate to give ions, they act as electrolytes and thus are electrically conductive. acids are sour in taste. example: hcl

bases are those substances which dissociates in water to give ions or accepts ions. as they dissociate to give ions, they act as electrolytes and thus are electrically conductive. bases are bitter in taste. bases are slippery in nature.