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.
It is A)The smallest unit of Ne
What are the temperatures
Empirical formula is the simplest ratio of components making up a compound.
The percentage composition of each element has been given
therefore the mass present of each element in 100 g of compound is
B N H
mass 40.28 g 52.20 g 7.53 g
number of moles
40.28 g / 11 g/mol 52.20 g / 14 g/mol 7.53 g / 1 g/mol
= 3.662 mol = 3.729 mol = 7.53 mol
divide the number of moles by the least number of moles, that is 3.662
3.662 / 3.662 3.729 / 3.662 7.53 / 3.662
= 1.000 = 1.018 = 2.056
the ratio of the elements after rounding off to the nearest whole number is
B : N : H = 1 : 1 : 2
therefore empirical formula for the compound is B₁N₁H₂
that can be written as BNH₂