There are two naturally occurring isotopes of boron determine the abundance of each isotope

Answer : The correct answer is:

The abundance of isotopes can be found out using following formula :

atomic mass of element = ( abundance of isotope 1 * mass of isotope 1 ) + ( abundance of isotope 2 * mass of isotope 2 ) +

Boron ( mass = 10.811 u ) has two isotopes : ¹⁰B ( mass = 10.0129 u ) and ¹¹ B ( mass = 11.0093 u )

Sum of abundance of all isotopes = 100%

Lets assume abundance of isotope ¹¹B = x % = 0.01 x , so abundance of isotope ¹⁰B = 100 % - x% = 100/100 - x/100 = 1 - 0.01 x

Plugging values in formula :

10.811 = ( 0.01 x * 11.0093) + [( 1 - 0.01 x) * 10.0129 ]

10.811 = (0.110093 x ) + ( 10.0129 - 0.100129 x )

10.811 - 10.0129 = 0.110093 x - 0.100129 x

0.7981 = 0.009964 x

Dividing both side by 0.00964

0.7981 / 0.009964 = 0.00964 /0.00964 x

x = 82.7 %

Abundance of of isotope ¹¹ B = 82.7 %

Abundance of isotope ¹⁰ B = 100% - x % = 100 %- 82.7 % = 17.3 %

the density of gold = 19.3 g/cm3

the relation between density and mass and volume is

density = mass / volume

given :

mass of gold = 13 g

density = 13g / volume = 19.3

volume = 13 /19.3 = 0.67 cm^3

thus volume of gold nugget = 0.67 cm^3

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explanation: