Question

Calculate the boiling point of a solution prepared by adding 15 gram of nacl to 250 gram of water

Answer 1

Answer:The boiling point of a solution prepared by adding 15 gram of nacl to 250 gram of water 373.53 K.

Explanation:

Mass of NaCl = 15 g

Weight of the solvent = 250 g

Boiling point of water =$T_o$ 373 K

boiling point of solution = T

K_b = 0.52 K kg/mol

$\Delta T=K_b\times molality=0.52\times\frac{\text{mass of NaCl}}{\text{molar mass of NaCl}\times \text{weight of the solvent(kg)}}$

$\Delta T=0.52\times\frac{15 g\times 1000}{58.5 g/mol\times 250}=0.53 K$

$\Delta T=0.53 K=T-T_o=T-373 K$

$T= 373.53 K$

Answer 2

Mass of solute (NaCl), Wb = 15 g

Mass of solvent (water), Wa = 250 g

Molar mass of NaCl, Mb = 58.44 g/mol

Kb for water = 0.512 K kg/mol

we have to calculate the boiling point of solution, Tb = ?

we know that boiling point of pure water, Tb° = 373 K

because,

∆Tb = Kb * m

∆Tb = Kb * Wb * 1000

Wa * Mb

= 0.512 * 15 * 1000

58.44 * 250

=> ∆Tb = 0.53 K

Now,

∆Tb = Tb - Tb°

Tb = Tb° + ∆Tb

= 373 + 0.53

=> Tb = 373.53 K

Hence, boiling point of solution is 373.53 K