CBSE Class 11 Chemistry Chapter 1: Some Basic Concepts of Chemistry NCERT Solutions

Solution: To determine accuracy and precision, we first calculate the average of the readings for each student and compare them to the correct value (3.0 g). Accuracy is the closeness of the average measurement to the true value, while precision is the closeness of individual measurements to each other.

For Student A:

Average reading = (2.99 + 3.01) / 2 = 3.00 g

The average reading (3.00 g) is very close to the correct reading (3.0 g), indicating high accuracy. The individual readings (2.99 g and 3.01 g) are also very close to each other (differ by 0.02 g), indicating high precision.

For Student B:

Average reading = (2.95 + 3.05) / 2 = 3.00 g

The average reading (3.00 g) is also very close to the correct reading (3.0 g), indicating high accuracy. However, the individual readings (2.95 g and 3.05 g) differ by 0.10 g, which is a larger spread, indicating lower precision compared to Student A.

Therefore, the results of student A are both precise and accurate.

Solution: We need to convert a temperature from the Fahrenheit scale to the Celsius scale. The relationship between Fahrenheit (°F) and Celsius (°C) is given by the formula:

$$^\{\backslash circ\}F\; =\; \backslash frac\{9\}\{5\}t^\{\backslash circ\}C\; +\; 32$$

To find the temperature in Celsius, we rearrange the formula:

$$t^\{\backslash circ\}C\; =\; \backslash frac\{5\}\{9\}(^\{\backslash circ\}F\; -\; 32)$$

Now, substitute the given Fahrenheit temperature (200°F) into the formula:

$$t^\{\backslash circ\}C\; =\; \backslash frac\{5\}\{9\}(200\; -\; 32)$$

$$t^\{\backslash circ\}C\; =\; \backslash frac\{5\}\{9\}(168)$$

$$t^\{\backslash circ\}C\; =\; \backslash frac\{840\}\{9\}$$

$$t^\{\backslash circ\}C\; =\; 93.33...^\{\backslash circ\}C$$

Rounding to one decimal place as often done in such conversions, we get 93.3 °C.

Thus, 200°F is equal to 93.3 °C.

Solution: Molarity is defined as the number of moles of solute dissolved per liter of solution. We are given the mass of the solute (NaCl) and the volume of the solution.

First, calculate the molar mass of NaCl. The atomic mass of Na is approximately 23 g/mol, and the atomic mass of Cl is approximately 35.5 g/mol. So, the molar mass of NaCl is $23\; +\; 35.5\; =\; 58.5$ g/mol.

Next, calculate the number of moles of NaCl:

$Moles\; \backslash \; of\; \backslash \; NaCl\; =\; \backslash frac\{Mass\; \backslash \; of\; \backslash \; NaCl\}\{Molar\; \backslash \; mass\; \backslash \; of\; \backslash \; NaCl\}$

$Moles\; \backslash \; of\; \backslash \; NaCl\; =\; \backslash frac\{5.85\; \backslash \; g\}\{58.5\; \backslash \; g/mol\}\; =\; 0.1\; \backslash \; mol$

The volume of the solution is given as 500 mL. To calculate molarity, we need the volume in liters:

$Volume\; \backslash \; in\; \backslash \; Liters\; =\; \backslash frac\{500\; \backslash \; mL\}\{1000\; \backslash \; mL/L\}\; =\; 0.5\; \backslash \; L$

Now, calculate the molarity:

$$Molarity\; =\; \backslash frac\{Moles\; \backslash \; of\; \backslash \; solute\}\{Volume\; \backslash \; of\; \backslash \; solution\; \backslash \; in\; \backslash \; Liters\}$$

$$Molarity\; =\; \backslash frac\{0.1\; \backslash \; mol\}\{0.5\; \backslash \; L\}\; =\; 0.2\; \backslash \; mol/L$$

Alternatively, using the formula Molarity = (weight × 1000) / (molecular weight × volume in mL):

$$Molarity\; =\; \backslash frac\{5.85\; \backslash \; \backslash times\; \backslash \; 1000\}\{58.5\; \backslash \; \backslash times\; \backslash \; 500\}\; =\; \backslash frac\{5850\}\{29250\}\; =\; 0.2\; \backslash \; mol/L$$

Therefore, the molarity of the solution is 0.2 mol L⁻¹.

Solution: This problem involves the dilution of a solution, where a stock solution is mixed with more solvent to decrease its concentration. We can use the dilution formula, which states that the initial molarity times the initial volume equals the final molarity times the final volume.

The formula is: $M\_1V\_1\; =\; M\_2V\_2$

Where:

• $M\_1$ = Initial Molarity = 5 M
• $V\_1$ = Initial Volume = 500 mL
• $M\_2$ = Final Molarity (what we need to find)
• $V\_2$ = Final Volume = 1500 mL

We need to solve for $M\_2$:

$$M\_2\; =\; \backslash frac\{M\_1V\_1\}\{V\_2\}$$

Substitute the given values into the formula:

$$M\_2\; =\; \backslash frac\{(5\; \backslash \; M)\; \backslash times\; (500\; \backslash \; mL)\}\{(1500\; \backslash \; mL)\}$$

$$M\_2\; =\; \backslash frac\{2500\; \backslash \; M\; \backslash cdot\; mL\}\{1500\; \backslash \; mL\}$$

$$M\_2\; =\; \backslash frac\{25\}\{15\}\; \backslash \; M\; =\; \backslash frac\{5\}\{3\}\; \backslash \; M$$

Converting the fraction to a decimal:

$M\_2\; \backslash approx\; 1.666...\; \backslash \; M$

Rounding to two decimal places, the final molarity is approximately 1.67 M. However, based on the provided options, 1.66 M is the closest and likely intended answer.

Thus, the molarity of the diluted solution is approximately 1.66 M.