When 2 g of a non-volatile solute was dissolved in 90 g of benzene the boiling point of benzene is raised by 0.88 K. Which of the following may be the solute? (Kb. for benzene .=2.53 K kg mol -1)

Question

When 2 g of a non-volatile solute was dissolved in 90 g of benzene the boiling point of benzene is raised by 0.88 K. Which of the following may be the solute? (Kb. for benzene .=2.53 K kg mol -1) (A) CO ( NH 2)2 (B) C 6 H 12 O 6 (C) NaCl (D) None of these.

Tardigrade Admin · Accepted Answer

wA=2 g , wB=90 g , MB=78 Δ Tb=0.88, Kb=2.53 K kg mol -1 Δ Tb=Kb × m 0.88=2.53 × (2 / mA/90) × 1000 ⇒ 0.88=(2.53 × 2/90 × mA) × 1000 mA=(2.53 × 2 × 1000/90 × 0.88)=63.8 ≈ 64 Molecular Mass of CO ( NH 2)2 ⇒ 60 Molecular Mass of C 6 H 12 O 6 ⇒ 180 Molecular Mass of NaCl ⇒ 58.5

Q. When $2 g$ of a non-volatile solute was dissolved in $90 g$ of benzene the boiling point of benzene is raised by $0.88 K$ . Which of the following may be the solute? $(K_{b}$ for benzene $= 2.53 K k g m o l^{- 1})$

$CO (N H_{2})_{2}$

$C_{6} H_{12} O_{6}$

$N a Cl$

None of these.

Q. When 2g of a non-volatile solute was dissolved in 90g of benzene the boiling point of benzene is raised by 0.88K. Which of the following may be the solute? (Kb​ for benzene =2.53Kkgmol−1)

CO(NH2​)2​

C6​H12​O6​

NaCl

None of these.

wA​=2g,wB​=90g,MB​=78 ΔTb​=0.88,Kb​=2.53Kkgmol−1 ΔTb​=Kb​×m 0.88=2.53×902/mA​​×1000 ⇒0.88=90×mA​2.53×2​×1000 mA​=90×0.882.53×2×1000​=63.8≈64 Molecular Mass of CO(NH2​)2​⇒60 Molecular Mass of C6​H12​O6​⇒180 Molecular Mass of NaCl⇒58.5

Q. When $2 g$ of a non-volatile solute was dissolved in $90 g$ of benzene the boiling point of benzene is raised by $0.88 K$ . Which of the following may be the solute? $(K_{b}$ for benzene $= 2.53 K k g m o l^{- 1})$

$CO (N H_{2})_{2}$

$C_{6} H_{12} O_{6}$

$N a Cl$