(4)

$m=\frac{1000\times M}{1000\times d-\left(M{M}_w\right)}$

Here, M = molarity of the solution, d = density of solution, $M_w$ = molar mass of the solute.

$m=\frac{1000\times 1}{(1000\times 1.25)-(1\times 85)}=\frac{1000}{1165}=0.858m$

(2)

$\text{Molality }\left(m\right)=\frac{\text{Moles of solute}}{\text{Mass of solvent in kg}}$

Let amount of solvent be $x$ g.

$1=\frac{0.5}{{\displaystyle \frac{x}{1000}}}$

$x=500$ $\text{g}$

(1)

Molarity is a function of temperature as volume depends on temperature.

(3)

1 molal aqueous solution means 1 mole of solute is present in 1000 g of water.

$\therefore$   $x_{\text{solute}}=\frac{1}{1+\frac{1000}{18}}=\frac{1}{56.5}=0.0177$