Asolution is made by dissolving 21.5 grams of glucose (c6h12o6) in 255 grams of water. what is the freezing point depression of the solvent if the freezing point constant is -1.86 °c/m? show all of the work needed to solve this problem.

We shall start by finding out the moles of the solute.

Moles of solute can be calculated as :

Moles=

Now,molar mass of the solute (C₆H₁₂O₆) will be = 6×12.0107+12×1.008+6×16

=180.73 grams/mol

Therefore moles of glucose will be:

Moles=21.5 grams/180.73 grams mol⁻¹

Moles=0.119 moles.

Now in order to calculate the freezing point depression ΔTf , we need to find the molality.

Molality=

m=0.119 moles/0.255 kg

m=0.46 molal.

The freezing point depression can be calculated as under :

∴ ΔTf=Kf × m

ΔTf = -1.86 °C/m × 0.46 m = -0.8556 °C

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.