Question

the density of an aqueous solution containing 12.50 g K2SO4 in 100.00 g solution is 1.083g/mL. Calculate the molality.

Answer 1

Molar Mass of K2SO4
K= 39*2= 78 amu
S= 32*1 = 32 amu
O=16*4 = 64 amu
--------------------------
Molar Mass of K2SO4 = 78 + 32 + 64 = 174 g/mol

Data:
W (Molality) = ?
m1 (mass of the solute) = 12.50 g
m2 (mass of the solvent) = 100.00 g (solution) - 12.50 g (solute) = 87.50 g (Solvent)
MM (molar mass) = 174 g/mol

Formula:

`W = (1000* m_(1) )/( m_(2)*MM )`

Solving:

`W = (1000* m_(1) )/( m_(2)*MM )`

`W = (1000*12.50 )/(87.50*174 )`

`W = (12500)/(15225)`

`\boxed{\boxed{W \approx 0.821 molal}}`