A compound `H_(2)X` with molar mass of `80 g` is dissolved in a solvent having density of `0.4 g mL^(-1)`. Assuming no change in volume upon dissolution, the molality of a `3.2` molar solution is

Answer

8

Explanation

Given 3.2 M solution
`therefore` moles of solute = 3.2 mol
Consider 1 L Solution.
`therefore` volume of solvent = 1 L
`P_("solvent")=0.4 g.mL^(-1) therefore m_("solvent")=P xx V=400 g`
`therefore` molality `=(3.2 mol)/(0.4 kg)= 8` molal.