Question

A student prepared a stock solution by dissolving 10.0 g of KOH in enough water to make 150. mL of solution. She then took 15.0 mL of the stock solution and diluted it with enough water to make water to make 65.0 mL of a final solution. What is the concentration of KOH for the final solution

Answer 1

Answer: The concentration of KOH for the final solution is 0.275 M

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution.

$Molarity=\frac{n\times 1000}{V_s}$

where,

n = moles of solute

 $V_s$ = volume of solution in ml = 150 ml

moles of solute =$\frac{\text {given mass}}{\text {molar mass}}=\frac{10.0g}{56g/mol}=0.178moles$

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

$Molality=\frac{0.178\times 1000}{150}=1.19M$

According to the dilution law,

$C_1V_1=C_2V_2$

where,

$C_1$ = molarity of stock solution = 1.19 M

$V_1$ = volume of stock solution = 15.0 ml

$C_2$ = molarity of diluted solution = ?

$V_2$ = volume of diluted solution = 65.0 ml

Putting in the values we get:

$1.19\times 15.0=M_2\times 65.0$

$M_1=0.275M$

Therefore, the concentration of KOH for the final solution is 0.275 M