Question

Calculate molality of a solution containing 2.5g of ethanoic acid in 75g of benzene

Answer 1

molality=no. of moles of solute/weight of solvent
no. of moles of solute= 2.5/60=25/600
=1/24=0.04mol
molality=0.04×1000/75=0.53 mol/g

Answer 2

Solution:-

⭐Molar mass of C$_{2}$H$_{4}$O$_{2}$ :-

• (12 × 2) + (1 × 4) + (16 × 2)

• 60 g mol$^{-1}$

❥Moles of C$_{2}$H$_{4}$O$_{2}$ :-

• $\frac{2.5 \: g}{ 60~g~mol~^{-1}}$

• 0.0417 mol

*️⃣Mass of benzene in kg :-

• $\frac{75~g}{1000~g~kg~^-1}$

• 75 × 10 $^{-3}$ kg

(⁠◕Molality of C$_{2}$H$_{4}$O$_{2}$ :-

• $\frac{Moles~of~C~_{2}~H~_{4}~O~_{2}}{kg~of~benzene}$

• $\frac{0.0417~mol~~×~~1000~g~kg~^{-1}}{75~kg}$

• 0.556 mol kg $^{-1}$