Question

A 3.00 laqueous solution of koh contains 195g of koh

Answer 1

Answer:

$\begin{gathered} a)\text{ 3.48 moles} \\ b)\text{ 0.9 mol/kg} \\ c)\text{ 4.80 \%} \end{gathered}$

Explanation:

a) Here, we want to calculate the molarity of the solution

Mathematically, we have that as:

$molarity\text{ = }\frac{number\text{ of moles}}{volume}$

But:

$\begin{gathered} number\text{ of moles = }\frac{mass}{molar\text{ mass}} \\ \\ molar\text{ mass of KOH = 56 g/mol} \\ \\ number\text{ of moles = }\frac{195}{56}\text{ = 3.48 moles} \end{gathered}$

Thus, we have the molarity as:

$molarity\text{ = }\frac{3.48}{3}\text{ = 1.16 M}$

b) Molality

Mathematically:

$molality\text{ = }\frac{moles\text{ of solute}}{kg\text{ of solvent}}$

To get kg of solvent:

$\begin{gathered} mass\text{ = volume }\times\text{ density} \\ mass\text{ = 3000 ml }\times\text{ 1.29 = 3870 g} \\ \\ To\text{ kg, we divide by 1000 = 3.87 kg} \end{gathered}$

From (a) above, we have the number of moles of solute as 3.48 moles

Thus, we have the molailty as:

$molality\text{ = }\frac{3.48}{3.87\text{ }}\text{ = 0.9 mol/kg}$

c) Mass percent concentration of solution

$Mass\text{ Percent = }\frac{Mass\text{ of solute}}{Mass\text{ of solution}}\text{ }\times\text{ 100 \%}$

The mass of the solute is 195 g

The mass of the solvent = 3870 g

The mass of the solution is 195 g + 3870 g = 4065 g

Thus, we have the mass percent as:

$\frac{195}{4065}\times\text{ 100 \% = 4.80 \%}$