Using the mass of the proton 1.0073 amu and assuming its diameter is 1.0×10−15m, calculate the density of a proton in g/cm3.

Answer :

Explanation :  To convert amu i.e. atomic mass unit in grams we have the conversion factor as 1 amu = 

we know the mass of the proton is 1.0073 amu

So converting it into grams we have to multiply;

1.0073 amu X   =

Now, Volume = 1/6πd³ as diameter is given as converting it to cm will require to multiply with 100

∴ Volume  = 1/6π

Hence, volume =  

Therefore, Density = mass / volume

∴ Density =  

Therefore, Density will be .

The density of a proton in g / cm₃ = 3.19.10⁻¹⁵

Atoms are composed of 3 types of basic particles namely protons, electrons, and neutrons

the mass of the basic particle is expressed in units of atomic mass

This atomic unit uses the standard atomic mass, that is, the C-12 isotope

1 atom C-12 = 12 atomic mass units

1 atomic mass unit = 1/12 x mass 1 C-12 atom

1 unit of atomic mass = 1.66.10 ⁻²⁴ g

the basic particle mass in atomic mass units is:

electron mass = 9.11.10⁻²⁸ g

proton mass = 1.6726.10⁻²⁴ g

neutron mass = 1,675.10⁻²⁴ g

The proton mass is considered to be equal to 1 atomic mass unit and is also considered to be equal to 1 atomic mass unit

1 unit of atomic mass = 1.6726.10⁻²⁴ g = 1.6726.10⁻²⁷kg

Density is a quantity derived from the mass and volume

density is the ratio of mass per unit volume

The unit of density can be expressed in g / cm³ or kg / m3³

Density formula:

ρ = density

m = mass

v = volume

Assume protons as spheres

with a diameter of 1.0 × 10⁻¹⁵m = 1.0.10⁻¹³ cm

then the volume = 4/3 π r³ or 1/6πd³

proton volume = 1 / 6π(1.0 × 10⁻¹³)³

proton volume = 5.23.10⁻⁴⁰

density becomes:

density = mass: volume

density = 1.6726.10⁻²⁴: 5.23.10⁻⁴⁰

density = 3.19.10⁻¹⁵ g/cm³

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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