We’d have to be very careful because if we had our skeletons on the outside it’d be very easy to injure ourselves
Molar mass Li2CO3 = 73.89 g/mol
Molar mass Li = 6.94g/mol Li = 6.94*2 = 13.88g
% LI = 13.88/73.89*100 = 18.78% perfectly correct.
Answer:
1 mole of HCl or NaOH gives you 1 mole of H2O , then the number of moles in H2O is: [ 1÷1×1 ] = 1 mole.
Explanation:
heating is repeated to ensure all water molecules have evaporated
<u>Answer: </u>The molar mass of solute is 115 g/mol.
<u>Explanation:</u>
Elevation in the boiling point is defined as the difference between the boiling point of the solution and the boiling point of the pure solvent.
The expression for the calculation of elevation in boiling point is:

OR
......(1)
where,
Boiling point of pure solvent (benzene) = 
Boiling point of solution = 
i = Vant Hoff factor = 1 (for non-electrolytes)
= Boiling point elevation constant = 
= Given mass of solute = 10 g
= Molar mass of solute = ? g/mol
= Mass of solvent = 200 g
Putting values in equation 1, we get:

Hence, the molar mass of solute is 115 g/mol.