6) What is the molality of 0.4 moles of KCl in 1500 g of solvent?

Molality = 0.267 mol/Kg

Explanation:

Given data:

Moles of KCl = 0.4 mol

Mass of solvent = 1500 g

Molality of solution = ?

Solution:

Formula:

Molality = moles of solute / kg of solvent

Now we will convert the g into kg.

1500 g× 1kg/ 1000 g

1.5 kg

Now we will put the values in formula.

Molality = 0.4 mol/ 1.5 kg

Molality = 0.267 mol/Kg

a) 1.392 x 10^6 g/cm^3

b) 8.69 x 10^7 lb/ft^3

Explanation:

mass of the star m =  2.0 x 10^36 kg

radius of the star (assumed to be spherical) r = 7.0 x 10^5 km = 7.0 x 10^8 m

The density of substance ρ = mass/volume

The volume of the star = volume of a sphere =

==> V = = 1.437 x 10^27 m^3

density of the star ρ = (2.0 x 10^36)/(1.437 x 10^27) = 1.392 x 10^9 kg/m^3

in g/cm^3 = (1.392 x 10^9)/1000 = 1.392 x 10^6 g/cm^3

in lb/ft^3 =  (1.392 x 10^9)/16.018 = 8.69 x 10^7 lb/ft^3