The naturally occurring isotopes of Li are Li-6 of mass 6.015121 amu and Li-7 of mass 7.016003 amu. The atomic mass of Li is 6.9409 amu, the percent abundance can be calculated using the following relation.
Atomic mass=m(Li-6 )×%(Li-6 )+m(Li-7 )×%(Li-7 )
Let the percent abundance of Li-6 be X thus, that of Li-7 will be 1-X, putting the values,
Or,
Or,
X=0.075
Thus,
Thus, percent abundance of Li-6 is 0.075 or 7.5 % and that of Li-7 is 0.925 or 92.5%.
Answer: 0.30 M
Explanation:
According to the neutralization law,
where,
= molarity of
solution = ?
= volume of
solution = 24.6 ml
= molarity of
solution = 0.222 M
= volume of
solution = 33.0 ml
= valency of
= 21
= valency of
= 1
Therefore, the molarity of the HCI solution is 0.30 M
The depression in freezing point is related to molality of solution as follows:
Here, is freezing point depression constant, for cyclohexane it is equal to 20°C kg/mol.
The value of freezing-point depression is 2.5 °C, molality can be calculated as follows:
Molality is defined as number of moles of solute in 1 kg of solvent.
Here, solvent is cyclohexane, its volume is given 20.0 mL and density is 0.779 g/mol thus, mass of cyclohexane can be calculated as follows:
Converting this into kg,
Thus, 15.58 g will be 0.01558 kg.
Now, number of moles of unknown solute is related to its mass and molar mass as follows:
Putting the values of mass of solute which is 0.2436 or 0.0002436 kg
Now, expression for molality of solution is:
Putting all the values,
Or,
On rearranging,
Or,
Therefore, molar mass of unknown sample is 125 g/mol
