A chemist adds 215.0mL of a 6.0x10^−5/mmolL mercury(II) iodide HgI2 solution to a reaction flask. Calculate the micromoles of mercury(II) iodide the chemist has added to the flask. Round your answer to 2 significant digits.

The micromoles of mercury(II) iodide : 0.013 μ moles

Given

215.0mL of a 6.0x10⁻⁵mmol/L HgI₂

Required

micromoles of HgI₂

Solution

Molarity(M) = moles of solute per liters of solution

Can be formulated :

M = n : V

n = moles

V = volume of solution

V = 215 mL = 0.215 L

so moles of solution :

n = M x V

n = 6.10 mmol/L x 0.215 L

n = 1.312 . 10⁻⁵ mmol

mmol = 10³ micromol

so 1.312 mmol = 1.312.10⁻⁵ x 10³ = 0.01312 micromoles ⇒ 2 sif fig = 0.013 μ moles

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

personal, she chose to do it.