Question

An aqueous solution of 4.57 M H2SO4 has a density of 1.25 g/mL. Calculate the molality of this solution

Answer 1

Answer : The molality of solution is, 5.69 mole/L

Explanation :

The relation between the molarity, molality and the density of the solution is,

$d=M[\frac{1}{m}+\frac{M_b}{1000}]$

where,

d = density of solution  = 1.25 g/mL

m = molality of solution  = ?

M = molarity of solution  = 4.57 M

$M_b =\text{molar mass of solute }(H_2SO_4)$ = 98 g/mole

Now put all the given values in the above formula, we get

$1.25g/ml=4.57M[\frac{1}{m}+\frac{98g/mole}{1000}]$

$m=5.69mol/kg$

Therefore, the molality of solution is, 5.69 mole/L

Answer 2

Answer:

The molality is 5.7 molal

Explanation:

Step 1: Data given

Molarity of H2SO4 = 4.57 M ( = 4.57 mol/L)

Density of the solution = 1.25 g/mL

Molar mass of H2SO4 = 98.09 g/mol

Step 2: Calculate mass of solution

Suppose we have 1L (= 1000 mL) of solution

Mass of the solution = 1.25 g/mL * 1000 mL

Mass of the solution = 1250 grams

Step 3: Calculate mass H2SO4

In 1L of a 4.57 M solution we have 4.57

Mass H2SO4 = moles H2SO4 * molar mass H2SO4

Mass H2SO4 = 4.57 moles * 98.09 g/mol

Mass H2SO4 = 448.3 grams

Step 4: Calculate mass solvent

Mass solvent = mass solution - mass H2SO4

Mass solvent = 1250 grams - 448.3 grams

Mass solvent = 801.7 grams

Step 5: Calculate molality

Molality = moles H2SO4 / mass solvent

Molality = 4.57 moles / 0.8017 kg

Molality = 5.7 molal

The molality is 5.7 molal