Answer
By dissolving $457.3 \text{ g}$ $ Ba(NO_3)_2$ in $ 1 \text{ kg}$ of water we can prepare $1.75 m$ $ Ba(NO_3)_2 $solution
Work Step by Step
molality $= 1.75 m$
molar mass of $Ba(NO_3)_2 = 137.33+28.02+96.00=261.35\text{ g/mol}$
if solvent water mass $= 1\text{ kg}$
molality $= moles / solvent$ $ mass$
$1.75 = moles / 1$
$moles = 1.75$
$mass = 1.75 \times 261.35 \approx 457.3 \text{ g}$
By dissolving $457.3 \text{ g}$ $ Ba(NO_3)_2$ in $ 1 \text{ kg}$ of water we can prepare $1.75 m$ $ Ba(NO_3)_2 $solution
