Question

The mole fraction of ch3oh in an aqueous solution is 0.02 and density of solution 0.994 g cm–3. determine the molality and molarity.

Answer 1

the molarity is 1.105 (M)

Answer 2

Answer:  Molarity of solution will be, 9.52 mole/L and the molality of solution will be, 13.38 mole/kg

Solution : Given,

Molar mass of $CH_3OH$ = 32 g/mole

0.02 moles of $CH_3OH$ in 0.08 moles of $H_2O$

Mass of $CH_3OH$(solute) = $0.02moles\times 32g/mole=0.64g$

Mass of $H_2O$= $0.08moles\times 18g/mole=1.44g$

Mass of solution = Mass of solute + Mass of solvent = 0.64+1.44 =2.08 g

Now we have to calculate the volume of solution.

$Volume=\frac{Mass}{Density}=\frac{2.08g}{0.994g/ml}=2.1ml$  

Molarity : It is defined as the number of moles of solute present in one liter of solution.

Formula used :

$Molarity=\frac{n}{V_s}$

where,

n = moles of solute $CH_3OH$ = 0.02

$V_s$ = volume of solution in liter

= 0.0021 L

Now put all the given values in the formula of molarity, we get

$Molarity=\frac{0.02}{0.0021L}=9.52mole/L$

Therefore, the molarity of solution will be, 9.52 mole/L

Molality : It is defined as the number of moles of solute present per kg of solvent.

$Molality=\frac{n}{W_s}$

n = moles of solute $CH_3OH$ = 0.02

$W_s$ = weight of solvent in kg= 0.00144 kg

$Molality=\frac{0.02}{0.00144kg}=13.88mole/kg$

Therefore, the molality of solution will be, 13.38 mole/kg.