The boiling point of a glucose solution containing 12 g of glucose in 100 g of water is 100. 34° C . Boiling point of water is 100° C. The molal elevation constant of water is

Question

The boiling point of a glucose solution containing 12 g of glucose in 100 g of water is 100. 34° C . Boiling point of water is 100° C. The molal elevation constant of water is (A) 0.51 ° C/Molal (B) 51 ° C/Molal (C) 5.1 ° C/Molal (D) None of these

Tardigrade Admin · Accepted Answer

Δ Tb=100.34-100=0.34° C Δ Tb=(kb × w2 × 1000/M2 × w1) kb=(Δ Tb × M2 × w1/w2 × 1000) kb=(0.34 × 180 × 100/12 × 1000)=0.51

Q. The boiling point of a glucose solution containing 12 g of glucose in 100 g of water is 100. $3 4^{\circ} C$ . Boiling point of water is $10 0^{\circ}$ C. The molal elevation constant of water is

0.51 $^{\circ}$ C/Molal

51 $^{\circ}$ C/Molal

5.1 $^{\circ}$ C/Molal

None of these

$Δ T_{b} = 100.34 - 100 = 0.3 4^{\circ} C$
$Δ T_{b} = \frac{k _{b} \times w _{2} \times 1000}{M _{2} \times w _{1}}$
$k_{b} = \frac{Δ T _{b} \times M _{2} \times w _{1}}{w _{2} \times 1000}$
$k_{b} = \frac{0.34 \times 180 \times 100}{12 \times 1000} = 0.51$

Q. The boiling point of a glucose solution containing 12 g of glucose in 100 g of water is 100. 34∘C . Boiling point of water is 100∘ C. The molal elevation constant of water is

0.51 ∘ C/Molal

51 ∘ C/Molal

5.1 ∘ C/Molal