The boiling point of water (100° C ) becomes 100.52° C, if 3 grams of a non-volatile solute is dissolved in 200 ml of water. The moleular weight of solute is (if K b for water is =0.6 K / m )

Question

The boiling point of water (100° C ) becomes 100.52° C, if 3 grams of a non-volatile solute is dissolved in 200 ml of water. The moleular weight of solute is (if K b for water is =0.6 K / m ) (A) 17.3 g.mol -1 : (B) 15.4 g.mol -1 : (C) 12.2 g.mol -1 : (D) 20.4 g.mol -1 :

Tardigrade Admin · Accepted Answer

As deration in boilding point is given by: Δ T=Kb × molality Putting the various values, we get: 100.52-100.00=0.6 × (3 / M/200 / 1000) ⇒ M=17.3 g mol -1

Q. The boiling point of water $(10 0^{\circ} C)$ becomes $100.5 2^{\circ} C$ , if $3$ grams of a non-volatile solute is dissolved in $200 m l$ of water. The moleular weight of solute is (if $K_{b}$ for water is $= 0.6 K / m$ )

$17.3 g . m o l^{- 1}$ :

$15.4 g . m o l^{- 1}$ :

$12.2 g . m o l^{- 1}$ :

$20.4 g . m o l^{- 1}$ :

As deration in boilding point is given by:

$Δ T = K_{b} \times$ molality

Putting the various values, we get:

$100.52 - 100.00 = 0.6 \times \frac{3/ M}{200/1000}$

$\Rightarrow M = 17.3 g m o l^{- 1}$

Q. The boiling point of water (100∘C) becomes 100.52∘C, if 3 grams of a non-volatile solute is dissolved in 200ml of water. The moleular weight of solute is (if Kb​ for water is =0.6K/m )

17.3g.mol−1 :

15.4g.mol−1 :

12.2g.mol−1 :

20.4g.mol−1 :