What mass of Na2SO4 must be dissolved in 104 grams of water to lower the freezing point by 2.50 °C? The freezing point depression constant, Kfp, of water is –1.86 °C/m. Assume t…

Question

asked Oct 5, 2020 60.4k views

1 Answer

Munkee · Answer 1 · 2020-10-09T23:47:05+0000

Answer: The mass of
that must be dissolved is 6.89 grams.

Step-by-step explanation:

Depression in freezing point is given by:

$\Delta T_f=i* K_f* m$

$\Delta T_f=2.50^0C$ = Depression in freezing point

i= vant hoff factor = 2.88

K_f = freezing point constant =
1.86^0C/m

m= molality

$\Delta T_f=i* K_f* \frac{\text{mass of solute}}{\text{molar mass of solute}* \text{weight of solvent in kg}}$

Weight of solvent (water)= 104 g = 0.104 kg

Molar mass of
Na_2SO_4 = 142 g/mol

Mass of
Na_2SO_4 added = ?

2.50=2.88* 1.86* (x)/(142* 0.104)

x=6.89g

The mass of
that must be dissolved is 6.89 grams.