<u>Answer:</u> The boiling point elevation of salt water solution is 0.55°C
<u>Explanation:</u>
To calculate the mass of water, we use the equation:
![\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}](https://tex.z-dn.net/?f=%5Ctext%7BDensity%20of%20substance%7D%3D%5Cfrac%7B%5Ctext%7BMass%20of%20substance%7D%7D%7B%5Ctext%7BVolume%20of%20substance%7D%7D)
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](https://tex.z-dn.net/?f=1g%2FmL%3D%5Cfrac%7B%5Ctext%7BMass%20of%20water%7D%7D%7B122mL%7D%5C%5C%5C%5C%5Ctext%7BMass%20of%20water%7D%3D%281g%2FmL%5Ctimes%20122mL%29%3D122g)
To calculate the elevation in boiling point, we use the equation:
![\Delta T_b=iK_bm](https://tex.z-dn.net/?f=%5CDelta%20T_b%3DiK_bm)
Or,
![\Delta T_b=i\times K_b\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}](https://tex.z-dn.net/?f=%5CDelta%20T_b%3Di%5Ctimes%20K_b%5Ctimes%20%5Cfrac%7Bm_%7Bsolute%7D%5Ctimes%201000%7D%7BM_%7Bsolute%7D%5Ctimes%20W_%7Bsolvent%7D%5Ctext%7B%20%28in%20grams%29%7D%7D)
where,
= ?
i = Vant hoff factor = 2 (For NaCl)
= molal boiling point elevation constant = 0.52°C/m.g
= Given mass of solute (NaCl) = 3.80 g
= Molar mass of solute (NaCl) = 58.55 g/mol
= 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](https://tex.z-dn.net/?f=%5CDelta%20T_b%3D2%5Ctimes%200.52%5EoC%2Fm%5Ctimes%20%5Cfrac%7B3.80%5Ctimes%201000%7D%7B58.55g%2Fmol%5Ctimes%20122%7D%5C%5C%5C%5C%5CDelta%20T_b%3D0.55%5EoC)
Hence, the boiling point elevation of salt water solution is 0.55°C