Question

If boiling point of urea solution is 100.18° C and K_b tor water is 0.512 K kg mol^-1, molality of solution is (Boiling point of water= 100° C)

विकल्प

• 0.6 mol kg^-1

• 0.25 mol kg^-1

• 0.35 mol kg^-1

• 0.45 mol kg^-1

Answer 1

0.35 mol kg^-1

Explanation:

Given,

Boiling point of solution = 100.18° C ⇒ 373.18 K

K_b = 0.512K kg mol^-1

Boiling point of water= 100° C ⇒ 373 K

`Δ"T"_"Boliling"` = 373.18 K - 373 K = 0.18 K

We know that,

ΔT = K_b × molarity

⇒ Molarity = `(Delta "T")/"K"_"b"`

`= (0.18 "K")/(0.512 "mol"^-1 "kg K")`

= 0.35 mol kg^-1