2.15 g of an unknown gas at 71 degrees c and 1.10 atm is stored in a 1.75 l flask. solve for the density and the molar mass of the gas.

Density :  0.01229 g/ml

Molar mass  : 31.525 g/mol

Some of the laws regarding gas, can apply to ideal gas (volume expansion does not occur when the gas is heated),:

So that the three laws can be combined into a single gas equation, the ideal gas equation

In general, the gas equation can be written

where

P = pressure, atm

V = volume, liter

n = number of moles

R = gas constant = 0.08206 L.atm / mol K

T = temperature, Kelvin

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 / m3

Density formula:

ρ = density

m = mass

v = volume

2.15 g of an unknown gas at 71 degrees C and 1.10 atm is stored in a 1.75 L flask.

so density = mass : volume

To find the molar mass, we first look for the number of moles with the ideal gas equation:

PV = nRT

T = 71 °C + 273 = 344 K

Which equation agrees with the ideal gas law

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Which law relates to the ideal gas law

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part 1: the half-life of the element is 22 years.

part 2: it will take 132.02 year for the sample to decay from 308.0 g to 4.8125 g.

explanation:

part 1: there are two ways to determine the half life time.

first way: the half life time can be determined directly from its definition as it is the time required for the sample to decay to its half concentration.

from the data the initial concentration was 45.0 g and decayed to its half value (22.5 g) in 22.0 years.

second way: the radio active decay is considered first order reaction so we can use the laws of first order reactions.

k = (1/t) ln (a/a-x), where k is the rate constant of the reaction,

t is the time of the reaction

a is the initial concentration of the reactant

a-x is the remaining concentration at time (t)

a = 45.0 g the initial concentration at t = 0 year.

at t = 22 year, a-x = 22.5 g

then k = (1/22 year) ln (45.0 g/ 22.5 g) = 0.0315 year-1.

the half-life time (t1/2) = ln 2/ k = 0.693/ k =0.693/ 0.0315 = 22 years.

part 2:

using the same law k = (1/t) ln (a/a-x),

using the given data; a = 308.0 g and a-x = 4.8125 g and the calculated k from the part 1 (k = 0.0315 year-1)

t = (1/k) ln (a/a-x) = (1/0.0315) ln (308.0/ 4.8125) = 132.02 years