In order to make a solution that freezes at -10 degrees celsius, how many grams of NaCl must you add to 3.0 kg?

When a compound is dissolved in water, the freezing point
of the solution decreases to below that of pure water. The decrease in temperature can
be found by using the following equation:

T = Kf * m * i,
where T is the change in the freezing point that is noticed, Kf is known as the
cryoscopic constant which is a constant for water irrespective of the solute and is
equal to 1.853 K*kg/mol, m is the molality of the solute and i is called the van't Hoff
factor which for NaCl is 2.

The freezing point has to be
decreased by 10 degree Celsius as the normal freezing point of pure water is 0 degree
Celsius. We need to determine the weight of salt that has to be added to 3 kg of
water.

10 = 1.853 * m *
2

=> m = 10/ (2*
1.853)

=> m = 2.698

The
required molality is 2.698 or we need to add 2.698 moles of NaCl to every kilogram of
water. 3 kg of water requires the addition of 3*2.698 = 8.094 moles. The mass of one
mole of NaCl is 58.443 g. The mass of 8.094 moles is 58.443*8.094 = 473.09
g

We need to add 473.09 g of NaCl to 3 kg of
water to reduce the freezing point to -10 degree
Celsius.