Chemistry, asked by surajraj8808, 1 year ago

18gram glucose dissolved in 500 gram water . Find the depression freezing point

Answers

Answered by mahakincsem
5

The freezing point to the solution is 272.8 K

Explanation:

To find out the freezing point, follow the procedure mentioned below

  • Weight of water = m1 = 500 g
  • Weight of glucose = m2 = 18 g
  • Molar mass of water = M1 =  18 g/mole
  • Molar mass of Glucose = M2 = 180 g/mole
  • Freezing point depression constant for water = kf = 1.86 Kk g/ mole

Now, applying the formula, we get

ΔT = Kf * m2 * 1000/ M2 * m1

ΔT  = 1.86 *18 *1000/ 180* 500

ΔT = 0.372

Then,

ΔT  = Tf° - Tf

Tf = Tf° -ΔT

Tf = 273.15- 0.372 = 272.8 K

Answered by Anonymous
5

Answer:

Explanation:

We know depression of freezing point is a colligative property.

∆T=Kf*m(where Kf is cryoscopic constant,m is morality)

Kf of water is 1.86°C/mm=No of moles/Mass of solvent in kg=(18/180)/(500/1000)=1/5

So, depression is 1.86*1/5=0.372°C

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