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

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)

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

0

explanation:

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