Question

A solution is prepared by dissolving 17.75 g sulfuric acid, h2so4, in enough water to make 100.0 ml of solution. if the density of the solution is 1.1094 g/ml, what is the molality?

Answer 1

Answer: The molality of solution is 1.94 m.

Explanation:

We are given:

Mass of sulfuric acid = 17.75 grams

Volume of solution = 100 mL

Density of solution = 1.1094 g/mL

To calculate the mass of solution, we use the equation:

$Density=\frac{Mass}{Volume}$

Putting values in above equation, we get:

$1.1094g/mL=\frac{\text{Mass of solution}}{100mL}$

$\text{Mass of solution}=110.94g$

Mass of solvent = Mass of solution - Mass of solute

Mass of solvent = 110.94 - 17.75 = 93.19 g

To calculate the molality of solution, we use the equation:

$Molality=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}$

Where,

$m_{solute}$ = Given mass of solute $(H_2SO_4)$ = 17.75 g

$M_{solute}$ = Molar mass of solute $(H_2SO_4)$ = 98 g/mol

$W_{solvent}$ = Mass of solvent = 93.19 g

Putting values in above equation, we get:

$\text{Molality of }H_2SO_4=\frac{17.75\times 1000}{98\times 93.19}$

$\text{Molality of }H_2SO_4=1.94m$

Hence, the molality of solution is 1.94 m.

Answer 2

When density = 1.1094 g / ml = 1.1094 Kg/ L &
the volume of solvent = 100m/1000=0.1 L
weight of solvent = denisty x volume = 1.1094 kg/L * 0.1 L = 0.111 Kg
∴ molality = no.of moles of solute / Kg of solvent 

we need to get the no.of moles of the solute = weight of solute/molar mass
                                                                           = 17.75 / 98 = 0.18
by substitution in molality equation:
∴ molality = 0.18/ 0.111 = 1.62 mol/kg