If 0.5 g of a solute (molar mass 100 g mol-1 ) in of solvent elevates the boiling point by 1K, the molar boiling point constant of the solvent is

Question

If 0.5 g of a solute (molar mass 100 g mol-1 ) in of solvent elevates the boiling point by 1K, the molar boiling point constant of the solvent is (A) 2 (B) 8 (C) 5 (D) 0.5

Tardigrade Admin · Accepted Answer

By using the relation Δ Tb=(kb× wB× 1000/MB× wA) Here, Δ Tb=1 K, wB=0.5g, wA=25g MB=100 g textmo textl-1 ∴ 1=(kb× 0.5× 1000/100× 25) kb=5

Q. If 0.5 g of a solute (molar mass 100 g $m o l^{- 1}$ ) in of solvent elevates the boiling point by 1K, the molar boiling point constant of the solvent is

2

8

5

0.5

By using the relation $Δ T_{b} = \frac{k _{b} \times w _{B} \times 1000}{M _{B} \times w _{A}}$ Here, $Δ T_{b} = 1 K,$ $w_{B} = 0.5 g,$ $w_{A} = 25 g$ $M_{B} = 100 g mo l^{- 1}$ $∴$ $1 = \frac{k _{b} \times 0.5 \times 1000}{100 \times 25}$ $k_{b} = 5$

Q. If 0.5 g of a solute (molar mass 100 g mol−1 ) in of solvent elevates the boiling point by 1K, the molar boiling point constant of the solvent is

2

8

5

0.5

By using the relation ΔTb​=MB​×wA​kb​×wB​×1000​ Here, ΔTb​=1K, wB​=0.5g, wA​=25g MB​=100gmol−1 ∴ 1=100×25kb​×0.5×1000​ kb​=5

Q. If 0.5 g of a solute (molar mass 100 g $m o l^{- 1}$ ) in of solvent elevates the boiling point by 1K, the molar boiling point constant of the solvent is

By using the relation $Δ T_{b} = \frac{k _{b} \times w _{B} \times 1000}{M _{B} \times w _{A}}$ Here, $Δ T_{b} = 1 K,$ $w_{B} = 0.5 g,$ $w_{A} = 25 g$ $M_{B} = 100 g mo l^{- 1}$ $∴$ $1 = \frac{k _{b} \times 0.5 \times 1000}{100 \times 25}$ $k_{b} = 5$