Question

Calculate the boiling point of a 4.0 m solution of sucrose in water. Sucrose is a nonelectrolyte. The normal boiling point of pure water is 100.0 degrees Celcius. Kb=0.52degrees celcius/m

Answer 1

Answer:

∆Tb= Kb × m

• 0.52× 4= 2.08
• Elevation in Boiling Point - 2.08
• Boiling Point - 100+2.08=102.08°C
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Answer 2

The boiling point of  a 4.0 m solution of sucrose in water is $\bold{102.048^{o}C}$.

Step by step explanation:

• The boiling point of the solution is calculate by the following formula.

        $\bold{T_{b}-T_{o}=K_{b}m}$..................(1)

•  From the given,

         molarity of the solution (m) = 4

         The boiling point of the water $\bold{T_{o}}$ = $100^{o}C$

          $K_{b}\,of\,H_{2}O=0.512\,c^{o}/m$

• Substitute the all given values in the equation (1)

       $T_{b}-100=0.512\times4$

        $T_{b}=102.048^{o}C$

• Therefore,The boiling point of  a 4.0 m solution of sucrose in water is $\bold{102.048^{o}C}$.