Answer : The average atomic mass of an element X is, 63.546 amu
Solution : Given,
Mass of isotope X-63 = 62.9296 amu
% abundance of isotope X-63 = 69.15% = 0.6915
Mass of isotope X-64 = 64.9278 amu
% abundance of isotope X-64 = 30.85% = 0.3085
Formula used for average atomic mass of an element X :
![\text{ Average atomic mass of an element X}=\sum[(62.9296\times0.6915)+(64.9278\times 0.3085)]](https://tex.z-dn.net/?f=%5Ctext%7B%20Average%20atomic%20mass%20of%20an%20element%20X%7D%3D%5Csum%5B%2862.9296%5Ctimes0.6915%29%2B%2864.9278%5Ctimes%200.3085%29%5D)
Therefore, the average atomic mass of an element X is, 63.546 amu
Answer:
0.0011 M.
Explanation:
Molarity of a substance , is the number of moles present in a liter of solution .
M = n / V
M = molarity
V = volume of solution in liter ,
n = moles of solute ,
Moles is denoted by given mass divided by the molecular mass ,
Hence ,
n = w / m
n = moles ,
w = given mass ,
m = molecular mass .
From the question ,
w = given mass of KMnO₄ = 0.0908 g
and ,
V = 0.500 L
Since ,
we know ,
m = Molecular mass of KMnO₄ = 158 g/mol
using , the above formula to calculate the moles ,
n = w / m = 0.0908 g / 158 g/mol = 0.00057 mol
now , calculating the molarity as ,
M = n / V = 0.00057 mol / 0.500 L = 0.0011 M.
