The molality is 0.54 M when 1.34 moles of NaCl is present in 2.47 kg of solvent.
<u>Explanation:</u>
Molality is the measure of how much of amount of solute is dissolved in the solvent. So it is calculated as the ratio of moles of solute to the grams of solvent.

As in this case, the solute is NaCl and solvent is unknown. So the moles of solute is given as 1.34 moles and the mass of solvent is given as 2.47 kg.
Hence, 
Thus, the molality is 0.54 M when 1.34 moles of NaCl is present in 2.47 kg of solvent.
Uses of isotopes:
In agriculture,
C-14 is used to trace the path of photosynthesis.
In medicine,
Iodine -131 is used in the treatment of
goiter
The reaction produces 2.93 g H₂.
M_r: 133.34 2.016
2Al + 6HCl → 2AlCl₃ + 3H₂
<em>Moles of AlCl₃</em> = 129 g AlCl₃ × (1 mol AlCl₃/133.34 g AlCl₃) = 0.9675 mol AlCl₃
<em>Moles of H₂</em> = 0.9675 mol AlCl₃ × (3 mol H₂/2 mol AlCl₃) = 1.451 mol H₂
<em>Mass of H₂</em> = 1.451 mol H₂ × (2.016 g H₂/1 mol H₂) = 2.93 g H₂
No, the formation of dew is condensation, which is a physical change.