What is the boiling point of a solution made by adding 6.69 g of magnesium chloride to 250.0 g of water? . use the formula of the salt to obtain i. (kb for water is 0.512 oc/m)

100.432ºC

Explanation:

Ebullioscopy is the elevation of the boiling point of a solvent that has a solute nonvolatile dissolved. The change in the temperature may be calculated by Raoult's law:

ΔT = Kb.W.i

Where ΔT is the temperature change, Kb is the ebullioscopy constant, W is the molality and i is the Van't Hoff factor, which determinates the particles that affect the proper.

The molality is:

W = m1/(M1xm2)

Where m1 is the solute mass (in g), M1 is the molar mass of the solute, and m2 is the mass of the solvent (in kg).

The Van't Hoff factor is the number of final particles divided by the number of initial particles. Magnesium is from froup 2, so it forms the cation Mg⁺², and chlorine ins from group 7 and forms the anion Cl⁻, the salt is MgCl₂ and dissociates:

MgCl₂  → Mg⁺²(aq) + 2Cl⁻(aq)

So, it has 1 particle in initial, and 3 in final (1  Mg⁺² and 2 Cl⁻). So:

i = 3/1 = 3.

The molar mass of MgCl₂ is: 24.3 + 2x35.5 = 95.3 g/mol, m1 = 6.69g, m2 = 0.250 kg, so:

W = 6.69/(95.3x0.250)

W = 0.281 m

Then:

ΔT = 0.512x0.281x3

ΔT = 0.432ºC

The normal boling point of water is 100ºC, so

T - 100 = 0.432

T = 100.432ºC

answer: the enthalpy of the overall chemical reaction is +87.88 kj or +88 kj.

according to hess's law, the overall enthalpy of a reaction is the difference between the sum of enthalpy values of all the products and the sum of enthalpy values of all the reactants.

δh(reaction) = ∑nδh(products) - ∑nδh(reactants)

δh(reaction) = [δh +δh] - [δh]

δh(reaction) = [1(-287.02) + 1(0) ] - [1(-374.9)] kj = +87.88 kj

here, δh (pcl3) = -287.02 kj, δh(cl2) = 0, δh(pcl5) = -374.9 kj

therefore, enthalpy of the overall reaction is +87.88 kj or +88 kj.

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol