NCERT Solutions for Class 12 Chemistry Chapter 2 Solutions

NCERT for Class 12 Chemistry Chapter 2 Solution – Free PDF Download
NCERT Solutions for Class 12 Chemistry Chapter 2 Solution – Free PDF Download give exact and sufficient Science information. The NCERT Answers for Class 12 Science can be used by the understudies to get ready for the board assessment and to address the activity inquiries of the Class 12 Science Part 2. Every one of the arrangements are made by the most recent CBSE Prospectus for 2023-24 and its rules.

These NCERT Answers for Class 12 Chemistry assist understudies with clearing every one of their questions connected with this part effortlessly. Further, understudies could likewise utilize this asset for planning notes and performing updates. Click the connection given beneath to download the NCERT Solutions for Class 12 Chemistry Chapter 2 PDF for free

Chemistry Class 12 NCERT Solutions Chapter 2 Solutions – Important Questions

Q :1 – If 22 g of benzene is dissolved in 122 g of carbon tetrachloride, determine the mass percentage of carbon tetrachloride (CCl4) and benzene (C6H6).

Ans:- To determine the mass percentages of benzene (C6H6) and carbon tetrachloride (CCl4) in the solution, you’ll need to use the following formulas:

Mass Percentage of Component X = (Mass of Component X / Total Mass of Solution) * 100

Mass Percentage of C6H6: $Mass Percentage of C6H6 = (22 g + 122 g 22 g) \times 100$
Mass Percentage of CCl4: $Mass Percentage of CCl4 = (22 g + 122 g 122 g) \times 100$

Calculating these values:

Mass Percentage of C6H6: $Mass Percentage of C6H6 = (144 22) \times 100 \approx 15.28%$
Mass Percentage of CCl4: $Mass Percentage of CCl4 = (144 122) \times 100 \approx 84.72%$

Therefore, the mass percentage of benzene (C6H6) is approximately 15.28%, and the mass percentage of carbon tetrachloride (CCl4) is approximately 84.72%.

Q 2.2) If benzene in a solution contains 30% by mass in carbon tetrachloride, calculate the mole fraction of benzene.

To calculate the mole fraction of benzene ( $X_{benzene}$ ), you can use the following formula:

$X_{benzene} = total moles in the solution moles of benzene$

The mass percent of benzene ( $% mass of benzene$ ) is given as 30%, which means that in a 100 g solution, 30 g is benzene.

Let’s assume we have 100 g of the solution. This means there are 30 g of benzene and (100 g – 30 g) = 70 g of carbon tetrachloride.

Now, we need to find the moles of benzene and carbon tetrachloride.

Find moles of benzene ( $n_{benzene}$ ): $n_{benzene} = molar mass of benzene mass of benzene$

The molar mass of benzene ( $C_{6} H_{6}$ ) is approximately 78.11 g/mol.

$n_{benzene} = 78.11 g/mol 30 g$

Find moles of carbon tetrachloride ( $n_{CCl 4}$ ): $n_{CCl 4} = molar mass of CCl ^{4} mass of CCl ^{4}$

The molar mass of carbon tetrachloride ( $CC l_{4}$ ) is approximately 153.82 g/mol.

$n_{CCl 4} = 153.82 g/mol 70 g$

Now, you can calculate the total moles ( $n_{total}$ ): $n_{total} = n_{benzene} + n_{CCl 4}$

Finally, calculate the mole fraction of benzene: $X_{benzene} = n ^{total} n ^{benzene}$

Substitute the values to get the final result.

Q 2.3) Determine the molarity of each of the solutions given below:

(a) 30 g of Co(NO)3. 6H2O in 4.3 L of solution.

(b) 30 mL of 0.5 M H2SO4 diluted to 500 mL

To determine the molarity ( $M$ ) of a solution, you can use the formula:

$M = volume of solution in liters moles of solute$

Let’s calculate the molarity for each solution:

(a) 30 g of Co(NO₃)₃·6H₂O in 4.3 L of solution:

Find moles of Co(NO₃)₃·6H₂O: $moles = molar mass mass$

The molar mass of Co(NO₃)₃·6H₂O can be calculated by adding the molar masses of cobalt (Co), nitrogen (N), oxygen (O), and hydrogen (H).

$Co(NO₃)₃\cdotp6H₂O=Molar mass of Co+3×(Molar mass of N+3×Molar mass of O)+6×Molar mass of H+6×Molar mass of O$

Now, substitute the values and calculate the moles.

Calculate Molarity ( $M$ ): $M = volume of solution in liters moles of solute$

Given that the volume of the solution is 4.3 L, you can substitute the values to find the molarity.

(b) 30 mL of 0.5 M H₂SO₄ diluted to 500 mL:

Determine moles of H₂SO₄: $moles = Molarity \times volume (in liters)$

Given that the initial volume is 30 mL and the molarity is 0.5 M, convert the volume to liters and calculate the moles.

Calculate Molarity of Diluted Solution: $M_{final} = volume of solution in liters moles of solute$

The final volume of the solution is 500 mL (0.5 L). Use this value to find the molarity of the diluted solution.

Please note that in the dilution process, the moles of solute remain constant, so you can set up an equation based on that principle to find the final molarity.

Q 2.4) To make 2.5 kg of 0.25 molar aqueous solution, determine the mass of urea (NH2CONH2) that is required.

To determine the mass of urea ( $NH_{2} CONH_{2}$ ) required to make a 0.25 molar aqueous solution, you can use the formula for molarity:

$Molarity = volume of solution in liters moles of solute$

The formula to calculate moles of solute is:

$moles = Molarity \times volume of solution (in liters)$

Given that you want to make a 0.25 molar solution with a total volume of 2.5 kg, first, convert the mass to volume using the density of water (assuming the solution is mainly water). The density of water is approximately 1 kg/L.

$Volume of solution (in liters) = density of water mass of solution$

Now, substitute the values into the equation to find the moles of urea:

$moles of urea = Molarity \times volume of solution (in liters)$

Finally, use the molar mass of urea to find the mass:

$mass of urea = moles of urea \times molar mass of urea$

The molar mass of urea ( $NH_{2} CONH_{2}$ ) can be calculated by adding the molar masses of nitrogen (N), hydrogen (H), carbon (C), and oxygen (O).

Once you calculate the mass of urea needed, you can proceed to prepare the solution.

NCERT Solutions for Class 12 Chemistry Chapter 2 Solutions

Chemistry Class 12 NCERT Solutions Chapter 2 Solutions – Important Questions

Q :1 – If 22 g of benzene is dissolved in 122 g of carbon tetrachloride, determine the mass percentage of carbon tetrachloride (CCl4) and benzene (C6H6).

Q 2.2) If benzene in a solution contains 30% by mass in carbon tetrachloride, calculate the mole fraction of benzene.

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