Question

How many grams of NaCl would need to be added to 1001 g of water to increase the boiling temperature of the solution by 1.500 °C? (Kb for water is 0.5100 °C/m)

Answer 1

Answer : The mass of NaCl needed would be, 86.1 grams.

Explanation :

Formula used for Elevation in boiling point :

$\Delta T_b=i\times k_b\times m$

or,

$\Delta T_b=i\times k_b\times \frac{w_2\times 1000}{M_2\times w_1}$

where,

$\Delta T_b$ = change in temperature = $1.500^oC$

$k_b$ = boiling point constant  = $0.5100^oC/m$

m = molality

i = Van't Hoff factor = 2 (for NaCl electrolyte)

$w_2$ = mass of solute (NaCl) = ?

$w_1$ = mass of solvent (water) = 1001 g

$M_2$ = molar mass of solute (NaCl) = 58.5 g/mol

Now put all the given values in the above formula, we get:

$1.500^oC=2\times (0.5100^oC/m)\times \frac{w_2\times 1000}{58.5g/mol\times (1001g)}$

$w_2=86.1g$

Therefore, the mass of NaCl needed would be, 86.1 grams.