Answer to Question #70744 in Chemistry for Desiree Berger

Question

1. Calculate the molarity of solution prepared by dissolving 80.0 g hydrogen fluoride in water to make 500.0 mL of solution; the more mass of HF is 20.01 g/mol.

2. A chemist needs to repair 2.50 mL of 0.100 M ammonia. How much concentrated (14.8 M) ammonia does he need?

3. Determine the freezing point of a solution containing 50.0 g of sucrose dissolved in 750. g water; round to the nearest 0.01°C. (Kf for water is 1.86°C/m, and a molar mass of sucrose is 342.3 g/mol.)

4. Explain why I cannot soft drink would be expected to fizz much more when opened at the peak of Mt. Everest (P ATM equals 260 TORR) then when open at sea level (P ATM equals 750 TORR)

Answer 1

1. CM = n/V
n (HF) = 80/20.01 = 4 mol
CM = 4/0.5 = 8 M

2. n1 = n2
0.100 · 2.50 = 14.8 · x
X = (0.100 · 2.50)/14.8 = 0.02 mL

3. Δt = i · Kf · m
t1= 0 ˚С, i=1 (sucrose is not ionic, so no dissociation), kf= 1.86 ˚С, m= mole of sucrose/kg of water
n = m/M
M (sucrose) = 342.3 g/mol
n (sucrose) = 50.0/342.3 = 0.15 mol
Cm (sucrose) = 0.15/0.75 = 0.19 m
Δt = 1 · 1.86 · 0.19 = 0.36 ˚C

4. According to Henry’s Law, the higher the pressure, the greater the solubility.
As the pressure at Everest is lower, soft drink would be expected to fizz much more when opened at the peak of Mt. Everest.