MHTCET 13 May 2023 Shift 1 Solved Paper

‌M2=‌

1000KbW2

∆TbW1

∴‌‌Kb=‌

M2∆TbW1

1000W2

‌Now2W2=‌ one gram mole ‌=M2g
‌W1=1‌kg=1000g
∴‌‌Kb=‌

M2×x×1000

1000×M2

=xK‌kg‌mol−1
One gram mole solute dissolved in 1‌kg solvent =1 molal solution
When concentration of solution is 1 molal, elevation in boiling point (∆Tb) is equal to molal elevation constant (Kb).
Therefore, Kb=xK‌kg‌mol−1