Post
-
Lab Questions
Comment on the accuracy of your line with respect to the
R
2
-
value. Give cee reason for each way the solutions were prepared that might throw points off the line.
If a CHEM
1

Question

Post
-
Lab Questions
Comment on the accuracy of your line with respect to the 
R
2
-
value. Give cee reason for each way the solutions were prepared that might throw points off the line.
If a CHEM 
1
0
1
 ﻿student makes 
2
0
0
 ﻿mL 
.
 ﻿of a cobalt solution that gives an absorbance of 
0
.
3
7
8
 
,
 ﻿using your graph, calculate how many grams of cobalt chloride bexahydrate did they weigh out?
Dr
.
 ﻿Dahm has a 
0
.
0
8
7
4
 ﻿M solution of nickel chloride. He pipettes 
1
0
 ﻿mL into a 
2
5
 ﻿mL 
.
 ﻿volumetric flask and adds water until the solution is at a final volume of 
2
5
 ﻿mL 
.
 ﻿Calculate his new molarity and determine what conductivity this solution should give using your graph.

VIPSolutions · Accepted Answer

# Lab Questions and Analysis

## Accuracy of the Line and $ R^2 $ Value

- The $ R2 $ indicates how well data fits the linear regression line.
- A high $ R^2 $ (close to 1) suggests good correlation; a low $ R^2 $ indicates more scatter or errors.
- **Potential reasons for points off the line:**
  - **Preparation errors:** Inaccurate pipetting, dilution mistakes, or inconsistent sample volumes.
  - **Instrumental errors:** Variability in absorbance readings or calibration issues.
  - **Sample impurities:** Contamination or impurities affecting absorbance.
  - **Environmental factors:** Temperature fluctuations affecting measurements.

---

## Calculation of Cobalt Chloride Dihydrate Mass

### Given Data:
- Volume used: $ 200\, \text{mL} $
- Absorbance: $ .378 $
- Graph relationship: $ \text{Absorbance} $ vs. $ \text{Concentration} $

### Step 1: Find concentration from the graph
- Using the graph, locate $ \text{Absorbance} = .378 $.
- Read off the corresponding concentration ($ C $) in mol/L.

*Assuming the graph indicates:*  
$$ C \approx .015\, \text{M} $$

### Step 2: Calculate moles of cobalt in solution
$$
\text{Moles} = C \times \text{Volume} = .015\, \text{mol/L} \times .200\, \text{L} = .003\, \text{mol}
$$

### Step 3: Convert moles to grams of cobalt chloride dihydrate ($ \mathrm{CoCl_2 \cdot 2H_2O} $)
- Molar mass of $ \mathrm{CoCl_2 \cdot 2H_2O} $:
  $$
  (58.93) + (2 \times 35.45) + (2 \times 18.02) = 237.93\, \text{g/mol}
  $$

- Grams of $ \mathrm{CoCl_2 \cdot 2H_2O} $:
  $$
  .003\, \text{mol} \times 237.93\, \text{g/mol} \approx \boxed{.7138\, \text{g}}
  $$

**Answer:** Approximately **.714 grams** of cobalt chloride dihydrate were weighed out.

---

## Dr. Dahm's Nickel Chloride Solution

### Step 1: Calculate new molarity
- Initial concentration: $ .0874\, \text{M} $
- Pipetted volume: $ 10\, \text{mL} = .010\, \text{L} $
- Final volume: $ 25\, \text{mL} = .025\, \text{L} $

Moles of nickel chloride initially:
$$
.0874\, \text{mol/L} \times .010\, \text{L} = 8.74 \times 10^{-4}\, \text{mol}
$$

Final concentration:
$$
C_{final} = \frac{\text{moles}}{\text{final volume}} = \frac{8.74 \times 10^{-4}\, \text{mol}}{.025\, \text{L}} = .035\, \text{M}
$$

### **Result:**
- **New molarity:** **.035 M**

---

## Conductivity Prediction Using the Graph

- Conductivity correlates with molarity; higher molarity generally results in higher conductivity.
- Using your graph, locate the conductivity value corresponding to $ .035\, \text{M} $.

*Assuming the graph indicates:*  
$$
\text{Conductivity} \approx 150\, \text{μS/cm}
$$

**Expected conductivity:** **~150 μS/cm**

---

# Summary

- The accuracy of your line depends on proper sample preparation and measurement consistency.
- The cobalt chloride dihydrate weighed was approximately **.714 g**.
- Dr. Dahm's solution has a molarity of **.035 M** and an estimated conductivity of **~150 μS/cm** based on the graph.

✨ VIPSolutions

Question:

Final Answer:

📢 Instructions:

Study Help, Fast Answers

✨ VIPSolutions

Question:

Final Answer:

📢 Instructions:

Related Answers

Study Help, Fast Answers