To solve this problem, we need to determine the volume (in liters) of a 0.0195 mmol/L potassium dichromate $(K_2Cr_2O_7)$ solution that contains 50.0 grams of potassium dichromate.

Step 1: Molar Mass of Potassium Dichromate

The molecular formula for potassium dichromate is $K_2Cr_2O_7$. We calculate its molar mass as follows:

• Potassium (K): $2 \times 39.10 = 78.20 \, \text{g/mol}$
• Chromium (Cr): $2 \times 52.00 = 104.00 \, \text{g/mol}$
• Oxygen (O): $7 \times 16.00 = 112.00 \, \text{g/mol}$

Thus, the molar mass of $K_2Cr_2O_7$ is: $78.20 + 104.00 + 112.00 = 294.20 \, \text{g/mol}$

Step 2: Convert Mass to Moles

Next, convert the 50.0 g of $K_2Cr_2O_7$ into moles: $\text{moles of } K_2Cr_2O_7 = \frac{50.0 \, \text{g}}{294.20 \, \text{g/mol}} = 0.170 \, \text{mol}$

Step 3: Convert to Millimoles

Since the concentration is given in millimoles per liter (mmol/L), convert the moles to millimoles: $0.170 \, \text{mol} = 170.0 \, \text{mmol}$

Step 4: Calculate Volume

Now, use the formula $C = \frac{n}{V}$ to calculate the volume: $V = \frac{n}{C}$ Where:

• $C = 0.0195 \, \text{mmol/L}$ (the concentration)
• $n = 170.0 \, \text{mmol}$

Thus: $V = \frac{170.0 \, \text{mmol}}{0.0195 \, \text{mmol/L}} = 8717.95 \, \text{L}$

Step 5: Significant Figures

The given data (50.0 g and 0.0195 mmol/L) has three significant figures, so the final answer should also have three significant figures. Therefore, the volume is: $\boxed{8.72 \times 10^3 \, \text{L}}$

Would you like a more detailed breakdown of any step or further clarifications?

Related Questions:

1. What is the process for converting between grams and moles in general?
2. How do significant figures affect calculations in chemistry?
3. Why is potassium dichromate used in molar concentration problems?
4. How do you convert between molarity and millimolarity?
5. What factors could affect the precision of such a calculation in a laboratory?

Tip:

Always make sure to match the number of significant figures in your final answer to the least precise measurement provided in the problem!