Question

An aqueous solution contains 28% w/w of KOH. If the density of the solution is 1.2 gm/ml, then calculate the :
1.) Molarity
2.)Molality
3.) %w/v
4.)Mole fraction of KOH

Answer 1

Answer:

Checkout the image

Explanation:

Answer 2

Answer:

1) The molarity measured is $6.25M$.

2) The molality measured is $6.94molal$.

3) The W/V % measured is $35\%$.

4) The mole fraction of $KOH$ measured is $11.11\%$.

Explanation:

Data given,

The concentration ( W/W % ) of $KOH$ = $28\%$

The density of the solution = $1.2g/ml$ = $1200g/L$

As given $28\%$, then $28g$ of $KOH$ in $100g$ of solution.

Therefore,

The mass of $KOH$ ( solute ) = $28g$

The mass of water ( solvent ) = $100- 28$ = $72g$ = $0.072kg$

As we know,

• The density of solution = $\frac{Mass of solution}{Volume of solution}$
• The volume of solution = $\frac{100}{1200}$ = $0.08L$ or $80ml$

1) Molarity

As we know,

• Molarity = $\frac{Number of moles}{Volume (L)}$

The molar mass of $KOH$ = $56g/mol$

Now, the number of moles of $KOH$, $n_{KOH}$ = $\frac{Given mass}{Molar mass}$ = $\frac{28}{56}$ = $0.5mol$

Thus,

• Molarity = $\frac{0.5}{0.08}$ = $6.25M$

2) Molality

As we know,

• Molality = $\frac{Number of moles (solute)}{Mass of solvent (kg)}$ = $\frac{0.5}{0.072}$ = $6.94molal$

3) W/V %

• W/V % = $\frac{mass of solute (g)}{volume of solution(ml)} \times 100$
• W/V % = $\frac{28}{80}\times 100$ = $35\%$

4) Mole fraction of $KOH$

As we know,

• Mole fraction of $KOH$ = $\frac{n_{KOH}}{n_{KOH} + n_{H_{2} O}}\times 100$

Now we have to find out the number of moles of water.

The molar mass of water = $18g/mol$

Therefore,

• $n_{H_{2}O }$ = $\frac{given mass}{molar mass}$ = $\frac{72}{18}$ = $4mol$

Thus, the mole fraction of $KOH$ = $\frac{0.5}{0.5+4}\times 100$ = $11.11\%$.