Question

Hii good morning all.................Q.Boiling point of water at 750 mm of Hg is 99.63°C.How much sucrose (moleculer mass 342u)is to be added to 500 g of water such that it boils at 100°C?Molal elevation boilig point constant,(Kb)for water is 0.52K kg/mol? please answer this question fast..........No spam plz❌​

Answer 1

Answer:

Explanation:

Refer the below attachment.

Answer 2

Explanation:

well....good afternoon

Here, elevation of boiling point ΔTb= (100 + 273) - (99.63 + 273)

= 0.37 K

Mass of water, wl = 500 g

Molar mass of sucrose (C12H22O11),

M2= 11 × 12 + 22 × 1 + 11 × 16 = 342 g mol - 1

Molal elevation constant, Kb= 0.52 K kg mol - 1

We know that:

∆Tb=kb*1000/m2*w1

=∆Tb*m2*w1/kb*1000

= (0.37 x 342 x 500) / (0.52 x 1000)

= 121.67 g (approximately)

Hence, 121.67 g of sucrose is to be added.