SolutionMedium

Question

The molal boiling point elevation constant of water is 0.513°C kg mol−1. When 0.1 mole of sugar is dissolved in 200 g of water, the solution boils under a pressure of 1 atm at

Options

A.100.513°C
B.102.565°C
C.100.256°C
D.101.025°C

Solution

$\Delta T_{b} = K_{b}.m = 0.513 \times \frac{0.1}{200/1000} = 0.2565^{o}C$

∴ B.P. of solution = 100.2565°C

Create a free account to view solution

View Solution Free
Topic: Solution·Practice all Solution questions

More Solution Questions

Of the following 0.10 m aqueous solutions, which one will exhibit the largest freezing point depression?...Glucose in blood can be quantitatively determined with :-...The Henry’s law constant for the solubility of N2 gas in water at 298 K is 1.0 × 105 atm. The mole fraction of N2 in air...Which of the following compound does not give test of sulphate ion in aqueous solution :-...Which of the following will cook food more easily?...