Question

Compute the boiling point elevation of a salt water solution that contains 3.80 g of NaCl dissolved in 122 mL of water. Enter the number of degrees celsius that the boiling point will be elevated.

Answer 1

Answer: The boiling point elevation of salt water solution is 0.55°C

Explanation:

To calculate the mass of water, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Density of water = 1 g/mL

Volume of water = 122 mL

Putting values in above equation, we get:

$1g/mL=\frac{\text{Mass of water}}{122mL}\\\\\text{Mass of water}=(1g/mL\times 122mL)=122g$

To calculate the elevation in boiling point, we use the equation:

$\Delta T_b=iK_bm$

Or,

$\Delta T_b=i\times K_b\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}$

where,

$\Delta T_b$ = ?

i = Vant hoff factor = 2 (For NaCl)

$K_b$ = molal boiling point elevation constant = 0.52°C/m.g

$m_{solute}$ = Given mass of solute (NaCl) = 3.80 g

$M_{solute}$ = Molar mass of solute (NaCl) = 58.55 g/mol

$W_{solvent}$ = Mass of solvent (water) = 122 g

Putting values in above equation, we get:

$\Delta T_b=2\times 0.52^oC/m\times \frac{3.80\times 1000}{58.55g/mol\times 122}\\\\\Delta T_b=0.55^oC$

Hence, the boiling point elevation of salt water solution is 0.55°C