Question

93.2 mL of a 2.03 M potassium fluoride (KF) solution

Answer 1

Answer:

$V_2=4.0132L$

Explanation:

Hello,

In this case, the original solution had a volume of 93.2 mL which in L is:

$V_1=93.2mL*\frac{1L}{1000mL}=0.0932L$

Now, since 3.92 L of water were added, the volume of the resulting solution will include the added volume of water as shown below:

$V_2=V_1+V_{H_2O}=3.92L+0.0932L\\\\V_2=4.0132L$

Moreover, the resulting molarity of the potassium fluoride diluted solution will be:

$M_2=\frac{M_1V_1}{V_2}=\frac{0.0932L*2.03M}{4.0132L}\\ \\M_2=0.0471M$

Best regards.

Answer 2

Answer:

1.98 M

Explanation:

Given data

• Initial volume (V₁): 93.2 mL

• Initial concentration (C₁): 2.03 M

• Volume of water added: 3.92 L

Step 1: Convert V₁ to liters

We will use the relationship 1 L = 1000 mL.

$93.2mL \times \frac{1L}{1000mL} = 0.0932 L$

Step 2: Calculate the final volume (V₂)

The final volume is the sum of the initial volume and the volume of water.

$V_2 = 0.0932L + 3.92 L = 4.01L$

Step 3: Calculate the final concentration (C₂)

We will use the dilution rule.

$C_1 \times V_1 = C_2 \times V_2\\C_2 = \frac{C_1 \times V_1}{V_2} = \frac{2.03 M \times 3.92L}{4.01L} = 1.98 M$