How many grams of table salt would have to dissolved in 145.0 g of water to lower the freezing point by 1.70 d

Answer 1

Let's set up our equation:

Drop in Freezing Point= kf(water)*2*molality of Salt

We multiply molality of salt by 2 because it yields two ions, Na+, Cl-, in solvent. It is also called the Vand't Hoff factor.

The Kf of water, a constant, is -1.86Celsius/mol

Molality is defined as moles solute/kilograms solvent. We have 145/1000=.145kg solvent (water): Now we just need to solve for moles.

-1.7= -1.86*2*molesNaCl/.145kg
-1.7(.145)=-3.72moles
molality=.06626moles of NaCl

Table Salt weight per mole is 23g/mol(Na)+35.5g/mol(Cl)=
58.5grams/mole

we have
.06626molesNaCl*58.5grams/mole
=3.88grams NaCl