Show the calculation of the molality of a solution made by dissolving 15.9 grams of Ca3(PO4)2 in 400 grams of water.

The molality of a solution : 0.128 m

Given

15.9 g of Ca3(PO4)2 in 400 grams of water.

Required

The molality

Solution

molality = mol solute / kg solvent

solute= Ca3(PO4)2(MW=Molecular Weight: 310.2 g/mol)

mol solute :

The molality :

first let us lay out the formula for potential

and kinetic energy:

pe = mgh

ke = (mv^2)/2

where: m=mass, v=velocity, g=gravitational

acceleration, h=height

calculating for mass using the known values of

pe:

340j = m (9.81 m/s^2) (50ft) (1m/3.28ft)

m = 2.27 kg

the law of conservation of energy states that

energy is neither created nor destroyed. therefore, the change in kinetic

energy from 50 ft to 35 ft would be equal to the change in potential energy

from 50 ft to 35 ft. however it would be opposite in signs since one is losing

while other is gaining.

ke(35ft) – ke(50ft) = - [ pe(35ft) – pe(50ft]

ke(50ft) = 0 since the brick is initially at rest

ke(35ft) = 340j - 2.27kg (9.81 m/s^2) (35ft)

(1m/3.28 ft)

ke(35ft)

= 102.38 j