Question

You have 500 mL of a 4.00 M solution of KNO3 from a previous experiment, but now you need 250 mL of a 0.500 M concentration of the same solution. How much of the stock solution will you need to make the dilute solution?

9 mL

20.0 mL

23.08 mL

9.09 mL

31 mL

What is the freezing point of a solution of 498mL of water (solute) dissolved in 2.50 L of ethanol (solvent), C2H5OH?
The density of C2H5OH is 0.789g/cm3. (Remember that water has a density of 1.0 g/cm3.)
-25.49°C

-142°C

-89.0°C

-73.3°C

-43.3°C

Answer 1

First question: 31 mL

Second question: T° freezing solution = -148.7°C

(It is logical to say -142°C if we look at the choices, but the value does not match)

Step-by-step explanation:

First question

Let's use dilution factor to solve the first question:

Diluited volume / Concentrated volume = M concentrated / M diluited

250 mL / Concentrated volume = 4 M / 0.5M

250 mL / Concentrated volume = 8

250 mL / 8 = Concentrated volume ⇒ 31 mL

We can also apply this formula

C₁ . V₁ = C₂ . V₂

4M . V₁ = 0.5M . 250 mL

V₁ = 31.25 mL

Second question → Colligative property about freezing point depression

ΔT frezzing = Kf . m . i

In this case i, is the Van't Hoff factor and it values 2

H₂O → H⁺ + OH⁻

m = molal (mol of solute in 1kg of solvent)

We have the volume of solvent, so with density we can know its mass.

Solvent mass = solvent volume / solvent density

Solvent mass = 2500 mL / 0.789 g/mL → 3168.5 g

(be careful with the units)

Solute mass = solute volume / solute density

498 g → solute mass, as the volume is 498 mL and density is 1 g/mL

Moles of water (mass / molar mass) ⇒ 498 g / 18 g/m = 27.6 moles

3168.5 g = 3.1685 kg

27.6 m / 3.1685 kg = 8.73 mol/kg

ΔT frezzing = T° freezing pure solvent - T° freezing solution

-114°C - (-T° freezing solution) = 1.99 °C/m . 8.73 m . 2

- T° freezing solution = 1.99 °C/m . 8.73 m . 2 + 114°C

T° freezing solution = -148.7°C