A solution of H _ 2 SO _ 4 is 31 . 4 % H _ 2 SO _ 4 by mass and has...

Given, Density= $1 . 25g / mL$,

Molar mass of ${H}_{2}{SO}_{4} = 98g{mol}^{- 1}$

Molarity is defined as no. of moles of solute present in one litre of solution. Thus,

Molarity= $\dfrac{No . ofmolesofsolute}{VolumeofsolutioninL}$

No. of moles= mass/ molar mass

$M = \dfrac{{n}_{\text{solute}}}{V} \times 1000$

$= \dfrac{\left(\dfrac{31 . 4}{98}\right)}{\left(\dfrac{100}{1 . 25}\right)} \times 1000$

$= 4 . 005 \approx 4$