The molecular weight of glucose, \text{c}_6 \text h_{12} \text o_6c 6 h 12 o 6 c, start subscript, 6, end subscript, h, start subscript, 12, end subscript, o, start subscript, 6, end subscript, is 180.16\,\dfrac{\text g}{\text{mol}}180.16 molg 180, point, 16, space, start fraction, g, divided by, m, o, l, end fraction. how many moles of glucose are in 19.1\,\text g19.1g19, point, 1, space, g of glucose? express the answer using 3 significant figures. \text{mol}mol

Explanation:

1) Data:

a) Chemical formula: C₆H₁₂O₆

b) Molecular weight, MM = 180.16 g/mol

c) mass, m = 19.1 g

c) n = ?

2) Formula:

3) Soltuion:

Substitute the data:

The significant figures start with the first non-zero digit after the decimal point, so they are 1, 0, and 6, i.e. three significant figures.

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

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