the boiling point of water is 100.00 °c at 1 atmosphere.

a student dissolves 11.11 grams of nickel(ii) iodide, nii2 (312.5 g/mol), in 272.7 grams of water.

what is the molality of the solution and boiling point?

The solution's boiling point is 100,2°C and its molality is 0,13 m

Explanation:

This is the colligative propertie about elevation of boiling point

ΔT = Kb . m . i where

ΔT is the difference between T° at boiling point of the solution - T° at boiling point of the solvent pure

Kb means ebulloscopic constant (0,52 °C.kg/m .- a known value for water)

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

i means theVan 't Hoff factor ( degree of dissociation for a compound)

IT HAS NO UNITS

NiI2 ---> Ni2+  +  2I-  (we have 1 Ni2+ and 2 I-), the i for this, is 3

The 11,11 g of the salt are in 272,2g of water but I need to know how many mass of the salt is in 1000 g of water (1000 g is 1 kg) so the rule of three is:

272,2g 11,11g

1000g (1000 g . 11,11g) / 272,2g = 40,81g

As the molar mas of NiI2 is 312.5 g/mol, the moles of salt are, mass/molar mass, 40,81g /312.5 g/mol = 0,130 moles

T° of b p sl - 100°C = 0,52 °C.kg/m . 0,130 m/kg . 3

T° of boiling point solution = (0,52 °C.kg/m . 0,130 m/kg . 3) + 100°C

T° of boiling point solution = 100,2°C

i believe the answer is a.