4.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)

5. A solution is made using 80.1 g of toluene (MM = 92.13 g/mol) and 80.0 g of benzene (MM = 78.11 g/mol). What is the molality of the toluene in the solution?

Answer :

(4) The mass of NaCl needed would be, 172.2 grams.

(5) The molality of toluene in the solution is, 10.9 mol/kg

Explanation :

Part 4:

for water =

Mass of water (solvent) = 1001 g  = 1.001 kg

Molar mass of NaCl = 58.5 g/mole

Formula used :  

where,

= change in boiling point

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

= boiling point constant for water

m = molality

Now put all the given values in this formula, we get

Therefore, the mass of NaCl needed would be, 172.2 grams.

Part 5:

Formula used :

Given:

Mass of toluene = 80.1 g

Mass of benzene = 80.0 g

Molar mass of toluene = 92.13 g/mol

Therefore, the molality of toluene in the solution is, 10.9 mol/kg

the temperature of the gas, in k, not celsius.

the answer is d.) potassium