Question

The rise in the boiling point of a solution containing 1.8 gms of glucose in

100 gms of a solvent is 0.1° C.Calculate the molal elevation constant of the

solvent. ( molar mass of glucose – 180)​

Answer 1

Using formula,

Using formula,ΔT

Using formula,ΔT b

Using formula,ΔT b

Using formula,ΔT b =K

Using formula,ΔT b =K b

Using formula,ΔT b =K b

Using formula,ΔT b =K b ×molality

Using formula,ΔT b =K b ×molality⟹0.1=K

Using formula,ΔT b =K b ×molality⟹0.1=K b

Using formula,ΔT b =K b ×molality⟹0.1=K b

Using formula,ΔT b =K b ×molality⟹0.1=K b ×

Using formula,ΔT b =K b ×molality⟹0.1=K b × 180×100

Using formula,ΔT b =K b ×molality⟹0.1=K b × 180×1001.8×1000

Using formula,ΔT b =K b ×molality⟹0.1=K b × 180×1001.8×1000

Using formula,ΔT b =K b ×molality⟹0.1=K b × 180×1001.8×1000

Using formula,ΔT b =K b ×molality⟹0.1=K b × 180×1001.8×1000 ⟹0.1=K

Using formula,ΔT b =K b ×molality⟹0.1=K b × 180×1001.8×1000 ⟹0.1=K b

Using formula,ΔT b =K b ×molality⟹0.1=K b × 180×1001.8×1000 ⟹0.1=K b

Using formula,ΔT b =K b ×molality⟹0.1=K b × 180×1001.8×1000 ⟹0.1=K b ×0.1

Using formula,ΔT b =K b ×molality⟹0.1=K b × 180×1001.8×1000 ⟹0.1=K b ×0.1⟹K

Using formula,ΔT b =K b ×molality⟹0.1=K b × 180×1001.8×1000 ⟹0.1=K b ×0.1⟹K b

Using formula,ΔT b =K b ×molality⟹0.1=K b × 180×1001.8×1000 ⟹0.1=K b ×0.1⟹K b

Using formula,ΔT b =K b ×molality⟹0.1=K b × 180×1001.8×1000 ⟹0.1=K b ×0.1⟹K b =1

Using formula,ΔT b =K b ×molality⟹0.1=K b × 180×1001.8×1000 ⟹0.1=K b ×0.1⟹K b =1 o

Using formula,ΔT b =K b ×molality⟹0.1=K b × 180×1001.8×1000 ⟹0.1=K b ×0.1⟹K b =1 o C m

Using formula,ΔT b =K b ×molality⟹0.1=K b × 180×1001.8×1000 ⟹0.1=K b ×0.1⟹K b =1 o C m −1

I HOPE THIS WILL HELP YOU MATE

Answer 2

Explanation:

your answer. click on this image and do your solution.