1) Molality of 1 M solution of NaNO₃ is
Density = 1.25 g/ml
A) 1.286m
B) 4.44m
C) 0.58m
D) None of these

Dear student,

Density of solution = 1.25 g/cm3
Therefore, mass of 1 L solution = volume x density = 1000 x 1.25 = 1250 g
Molar mass of NaNO3 = 85 g
As we have given 1 M solution of NaNO3, it means 1 mole of NaNO3 is present in 1 L volume. 
Mass of 1 mole of NaNO3 = 85 g
Mass of solvent = 1250 - 85 = 1165 g = 1.165 kg

Molality of NaNO3 = number of moles of solute / mass of solvent in kg
= 1 / 1.165
= 0.858 m

THe answer given in the options does not match. Hence, it will be none of these.

Regards
Regards