Gasoline and kerosene (specific gravity 0.820) are blended to obtain a mixture with a specific gravity of 0.770. calculate the volumetric ratio (volume of gasoline/volume of kerosene) of the two compounds in the mixture, assuming vblend = vgasoline + vkerosene.

The volumetric ratio is 0,71

Explanation:

Let's begin with the equation:

(1)

Where:

Db: Blend Density, Mb: Blend Mass and Vb: Blend Volume

And we know: (2)

Where:

Vg: Gasoline Volume and Vk: Kerosene Volume

Therefore replacing (2) into (1):

(3)

Where:

Dg: Gasoline Density and Dk: Kerosene Density

The specific gravity is defined as:

Therefore:

Where:

Dref: Reference Density

SGb: Blend Specific Gravity

SGg: Gasoline Specific Gravity (which is 0.7 approximately)

SGk: Kerosene Specific Gravity

Replacing these equations into (3) we get:

Replacing with the Specific Gravity data, we obtain:

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