A chemist working as a safety inspector finds an unmarked bottle in a lab cabinet. A note on the door of the cabinet says the cabinet is used to store bottles of pentane, ethanolamine, diethylamine, glycerol, and carbon tetrachloride. The chemist plans to try to identify the unknown liquid by measuring the density and comparing to known densities. First, from his collection of Material Safety Data Sheets (MSDS), the chemist finds the following information: liquid density pentane 0.63gmL, ethanolamine 1.0gmL, diethylamine 0.71gmL, glycerol 1.3gmL, carbon tetrachloride 1.6gmL Next, the chemist measures the volume of the unknown liquid as 1737.cm3 and the mass of the unknown liquid as 1.23kg . Calculate the density of the liquid. Be sure your answer has the correct number of significant digits.

0.71 g/mL

The liquid is diethylamine

Explanation:

Mass of the liquid = 1.23 kg of 1.23 * 10^3 g

Volume of the liquid = 1737.cm3

Let us remember that density is obtained as;

Density= mass/volume

Density= 1.23 * 10^3 g/1737cm^3

Density= 0.71 g/cm^3

Since we know that 1 mL=1 cm3 . So, 1 g/mL=1 g/cm3

So we can also write;

Density= 0.71 g/mL

The liquid is diethylamine

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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