G Sucrose (C12H22O11), a nonionic solute, dissolves in water (normal freezing/melting point 0.0°C) to form a solution. If some unknown mass of sucrose is dissolved in 150.g of water and this solution has a freezing point of -0.56°C, calculate the mass of sucrose dissolved.

15.4 g of sucrose

Explanation:

Formula to be applied for solving these question: colligative property of freezing point depression. → ΔT = Kf . m

ΔT = Freezing T° of pure solvent - Freezing T° of solution

Let's replace data given: 0°C - (-0.56°C) = 1.86 C/m . m

0.56°C / 1.86 m/°C = m → 0.301 mol/kg

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

Our mass of solvent is not 1kg, it is 150 g. Let's convert it from g to kg, to determine the moles of solute: 150 g. 1kg/1000g = 0.150 kg

0.301 mol/kg .  0.150kg = 0.045 moles.

We determine the mass of sucrose, by the molar mass:

0.045 mol . 342 g/1mol = 15.4 g

substituting a hydrogen atom with a halogen in a hydrocarbon:

- the single bond remains single. a is wrong.

- the bond with hydrogen is as strong as with a halogen. c is wrong

- a hydrocarbon is saturated if there is no double bond. we do not know that so d is wrong.

- carbon will always have four valence electrons. e is wrong.

by elimination, the correct answer is b. the boiling point of the new compound increases.