Question

Calculate the molar mass of the substance 1.3 g of which when dissolved in 169g of water gave the solution which will boil at 100.025°C at 1atm.(k^b=0.52Km^-1)

Answer 1

Explanation:

Given Calculate the molar mass of the substance 1.3 g of which when dissolved in 169g of water gave the solution which will boil at 100.025°C at 1 atm.(k^b=0.52K kgm^-1)

• Now we need to find the molar mass of the substance.
• So mass of substance Wb = 1.3 g
• Weight of water = Wa = 169 g
• So K b = 0.52 k kg m^-1
• So initial boiling point of water will be
• So Δ Tb = 100.025 – 100 = 0.025 degree Celsius
• Now we have molality
• So Δ Tb = Kb m
•              = Kb Wb / Mb x 1000 / Wa
• Or Mb = Kb x Wb x 1000 / ΔTb x Wa
•            = 0.52 x 1.3 x 1000 / 0.025 x 169
•           = 676 / 4.225
•          = 160 g / mol

Reference link will be

https://brainly.in/question/6047121

Answer 2

Answer:

boiling point of pure water is 100*C vich is t*b

Bp of solution is given=100.025

w2=1.3g

w1=169g

kb=0.52

putting these values in the formula..

M2=kbxW2x1000 divided by 0.025 x 169

here 0.025 id delta tb. which is obtained by subtracting tb-t*b

= 100.025-100=0.025

after solving ...

676 divided by 4.225 (calculation of values in formula used above)

we get the answer=160gmol-1