The boiling point of water is 100oC . What will be the boiling point of an aqueous solution containing 0.6 g of urea (molar mass =60 ) in 100 g of water?( Kb for water =0.52 K/m )

Question

The boiling point of water is 100oC . What will be the boiling point of an aqueous solution containing 0.6 g of urea (molar mass =60 ) in 100 g of water?( Kb for water =0.52 K/m ) (A)  100.052oC  (B)  101.052oC  (C)  120.52oC  (D) None of these

Tardigrade Admin · Accepted Answer

We know that, Δ Tb=(Kb× wB× 1000/MB× wA) Kb=0.52K m-1, wB=0.6 g, Mb=60 wA=100 g ∴ Δ Tb=(0.52× 0.6× 1000/60× 1000)=0.052oC Thus boiling point of the solution =100+0.052=100.052oC

Q. The boiling point of water is $100^{o} C$ . What will be the boiling point of an aqueous solution containing $0.6 g$ of urea (molar mass $= 60$ ) in $100 g$ of water?( $K_{b}$ for water $= 0.52 K / m$ )

$100.052^{o} C$

$101.052^{o} C$

$120.52^{o} C$

None of these

Q. The boiling point of water is 100oC . What will be the boiling point of an aqueous solution containing 0.6g of urea (molar mass =60 ) in 100g of water?( Kb​ for water =0.52K/m )

100.052oC

101.052oC

120.52oC

None of these

We know that, ΔTb​=MB​×wA​Kb​×wB​×1000​ Kb​=0.52Km−1,wB​=0.6g,Mb​=60 wA​=100g ∴ ΔTb​=60×10000.52×0.6×1000​=0.052oC Thus boiling point of the solution =100+0.052=100.052oC

Q. The boiling point of water is $100^{o} C$ . What will be the boiling point of an aqueous solution containing $0.6 g$ of urea (molar mass $= 60$ ) in $100 g$ of water?( $K_{b}$ for water $= 0.52 K / m$ )

$100.052^{o} C$

$101.052^{o} C$

$120.52^{o} C$