When 105. g of urea CH4N2O are dissolved in 650. g of a certain mystery liquid X, the freezing point of the solution is 2.90°C less than the freezing point of pure X. Calculate the mass of potassium bromide that must be dissolved in the same mass of X to produce the same depression in freezing point.

104 g of KBr

Explanation:

Formula por freezing point depression is:

ΔT = Kf . m . i

where ΔT is the Freezing T° of pure solvent - Freezing T° of solution

In the first case, ΔT = 2.90°C

We need to determine the Kf for X, to apply the formula again and determine the mass of KBr

m = molality (moles of solute in 1kg of solvent)

We convert 650 g to kg → 0.65 kg

We need to determine the moles of urea (solute)

105 g . 1 mol / 60 g = 1.75 mol

m for urea is: 1.75 mol /0.65 kg = 2.69 m

i = number of ions dissolved. Urea is an organic compound, so no ions.

i = 1

2.90°C = Kf . 2.69 m . 1

2.90°C / 2.69 m = 1.08 °C/m

In the next salt, KBr this is a ionic salt that can be dissociated

KBr →  K⁺  +  Br⁻

2 ions are generated, so i = 2

We replace again: 2.90°C = 1.08°C/m . m . 2

2.90°C / (1.08°C/m . 2) = m

molality = 1.34

Moles of solute in 1kg of solvent. Let's find out the moles in the same mass of X, we had before.

1.34 mol/kg . 0.65 kg = 0.873 moles

We convert mass to moles: 0.873 mol . 119 g /mol = 104 g

b..

is the correct one

explanation:

covalent bonds form as valence electrons are shared between two atoms. lewis structures and structural formulas are common ways of showing the covalent bonding in organic molecules. formal charge describes the changes in the number of valence electrons as an atom becomes bonded into a molecule.

2b2t

explanation:

2b2t