NUMERICAL PRACTICE

DAY 5
27.4.20

Good Morning Boys,
Today we will be practising the numericals related to the concept learnt in previous classes.

Learning outcome :
students will be able to

• Apply the formula
• interpret the question.
• Analyse the question.

Molarity formula with Density is given as

conversion table for reference :

1 cm3	1mL	1000 mm3
1 Litre	1000mL	1000 cm3
1 m3	106  cm3	1000 L
1 dm3	1000 cm3	1 L

NUMERICAL QUESTIONS :

Question 1
Calculate the molarity of $NaOH$ in the solution prepared by dissolving its 4 g in enough water to form 250 mL of the solution.
Solution
 Molarity M = Moles of solute(g)/ volume of solution (litres)
= (4 g / 40 g ) ×  (1000/250) =0.1 mol/0.250 L
= 0.4 mol/L
= 0.4 M
Question 2

20% (w/W) NaOH solution (density of solution= 1.2 g/cc). Calculate the Molarity?
Solution
20% (w/M) means 20 gm  in 100gm  of solution
Now Density of solution =1.2g/cc
Density = mass /volume
So Volume of Solution =100/1.2 cc
Now
M = W/M  ×  1000/(V ml)
M= 20/40 ×  1000/100 × 1.2 = 6  M

Question 3
Calculate the molarity of 2% (m/v) glucose sol.
Solution
2 gm Glucose in 100 ml sol.
Now Molar mass of glucose(C6 H12 O6) = 12×6+1 ×12+6
 = 72 + 12 + 96=180 gm
                
M = W/M  ×  1000/(V ml)
M = 2/180  ×1000/100
M = 1/9

Question 4
If 500 mL of a 5M solution is diluted to 1500 mL, what will be the molarity of the solution obtained?
Solution
M₁ V₁ = M₂ V₂
5 X 500= M_{2  x 1500}
M₂ =1.66 M

Question 5
If 4 g of NaOH dissolves in 36 g of H₂O, calculate the mole fraction of each component in the solution. Also, determine the molarity of solution (specific gravity of solution is 1g mL–1).
Solution
Mass of NaOH = 4 g
Number of moles of NaOH =4/40
= 0.1 mol

Mass of H2O = 36 g
Number of moles of  H₂O= 36/18 =2 mol

Mole fraction of water
=Number of moles of H₂O, /(No. of moles of water + No. of moles of NaOH)
=2/(2+0.1)  = 0.95

Mole fraction of NaOH =Number of moles of NaOH/ (No. of moles of NaOH + No. of moles of water)

= 0.1 /(2+0.1) = 0.047

Mass of solution = mass of water + mass of NaOH = 36 g + 4 g = 40 g
Volume of solution = 40 × 1 = 40 mL
(Since specific gravity of solution is = 1 g mL–1)
Molarity of solution
=Number of moles of solute/ Volume of solution in litre
=0.1 mol NaOH /.04 = 2.5 M
Question 6

How to prepare 0.2 M HCL?
Solution
Now As per Molarity Formula
For 0.2M , we need 0.2 moles of HCl in 1 Litre Solution
Now Molecular mass of HCl=36.5 g/mol
1 mole of HCl contain $1\times \mathrm{36.\; 5\; g\; of\; Hcl\; .\; so}$0.2 moles of HCl contain $0.2\times 36.5=7.3$g HCl
Therefore  7.3 g of HCl in a litre of solution gives 0.2 molar HCl solution

Question 7
An aqueous solution is prepared by diluting 3.30 mL acetone (d = 0.789 g/mL) with water to a final volume of 75.0 mL. The density of the solution is 0.993 g/mL. What is the molarity, molality and mole fraction of acetone in this solution?

Solution:

1) Preliminary calculations:
Density = mass /volume,
therefore

mass of acetone = (3.30 mL) x (0.789 g/mL) = 2.6037 g
moles of acetone: 2.6037 g / 58 g/mol = 0.0448 mol 
mass of solution: (75.0 mL) x (0.993 g/mL) = 74.475 g
mass of water in the solution: 74.475 g - 2.6037 g = 71.8713 g
moles of water: 71.8713 g / 18 g/mol = 3.9928 mol

2) Molarity: moles of solute / vol of solution (lt)

0.0448mol / 0.0750 L = 0.598 M

3) Molality: moles of solute/ mass of solvent (kg)

0.04483 mol / 0.0718713 kg = 0.624 m

4) Mole fraction: moles of one component/ total no of moles in solution

0.0448 mol / (0.0448 mol + 3.9928 mol) = 0.0110

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Question: 8

What is the weight percentage of urea solution in which 10 gm of urea is dissolved in 90 gm water.

Solution

Weight percentage of urea = (weight of urea/ weight of solution) 100

 = 10/(90+10)  100 = 10% urea solution (w/W)

QUESTION 9

A sulphuric acid solution containing 571.4 g of H₂SO₄ per litre of solution has a density of 1.329 g/cm³. Calculate the molality of H₂SO₄ in this solution

Solution:

1 L of solution = 1000 mL = 1000 cm^3.     1.329 g/cm³ times 1000 cm³ = 1329 g (the mass of the entire solution)
1329 g - 571.4 g = 757.6 g = 0.7576 kg (the mass of water in the solution)
mol of H₂SO₄    571.4 g / 98 g/mol = 5.8306  moles.
 molality (m) = 5.8306 mol / 0.7576 kg = 7.6906m

QUESTION 10
 Determine the mole fraction of CH3OH and H2O in a solution prepared by dissolving 5.5 g of alcohol in 40 g of H2O. Molecular mass of H2O is 18  u and Molecular mass  of CH3OH is 32.

Solution

Moles of CH3OH = 5.5 / 32 = 0.17 mole

Moles of H2O = 40 / 18 = 2.2 moles

Therefore, according to the equation

mole fraction of CH3OH = 0.17 / 2.22 + 0.17

mole fraction of CH3OH = 0.071
mole fraction of water  = 2.22/0.17+2.22 = 0.92

QUESTION 11

For the combustion of sucrose:

C₁₂H₂₂O₁₁ + 12O₂ ---> 12CO₂ + 11H₂O

there are 10.0 g of sucrose and 10.0 g of oxygen reacting. Which is the limiting reagent?

Solution 

1) Calculate moles of sucrose:

10.0 g / 342g/mol = 0.029 mol

2) Calculate moles of oxygen required to react with moles of sucrose:

From the coefficients, we see that 12 moles of oxygen are require for every one mole of sucrose. Therefore: for 0.029 mol  of sucrose the amount of oxygen  required would be 0.349 moles.

3) Determine limiting reagent:

Oxygen on hand ⇒ 10.0 g / 32g/mol = 0.3125 mol

since oxygen is present is less amount(0.3125) as compared to the requirement of oxygen as  0.349 moles 
therefore, oxygen is the limiting reagent.

  QUESTION 12 

Calculation of Percent Composition
Analysis of a 12.04-g sample of a liquid compound composed of carbon, hydrogen, and nitrogen showed it to contain 7.34 g C, 1.85 g H, and 2.85 g N. What is the percent composition of this compound?

Solution

$\%\text{C}=\frac{7.34\text{g C}}{12.04\text{g compound}}\times 100\%=61.0\%$
$\%\text{H}=\frac{1.85\text{g H}}{12.04\text{g compound}}\times 100\%=15.4\%$
$\%\text{N}=\frac{2.85\text{g N}}{12.04\text{g compound}}\times 100\%=23.7\%$

QUESTION 13 

Aspirin is a compound with the molecular formula C9H8O4. What is its percent composition?

Solution
 consider 1 mol of C9H8O4  SO  its molar mass is (180 g/mole), 

$\begin{array}{cc}\%\text{C}& \frac{9\text{mol C}\times \text{molar mass C}}{\text{molar mass}{\text{C}}_9{\text{H}}_{18}{\text{O}}_4}\times 100=\frac{9\times \mathrm{12\; g/mol}}{\mathrm{180g/mol}}\times 100=\frac{}{}\\ \%\text{C}& \mathrm{60\%}\text{C}\end{array}$

$\begin{array}{cc}\%\text{H}& \frac{8\text{mol H}\times \text{molar mass H}}{\text{molar mass}{\text{C}}_9{\text{H}}_{18}{\text{O}}_4}\times 100=\frac{8g/mol}{180.\text{g/mol}}\times 100=\frac{}{}\\ \%\text{H}& 4.44\%\text{H}\end{array}$

$\begin{array}{cc}\%\text{O}& \frac{4\text{mol O}\times \text{molar mass O}}{\text{molar mass}{\text{C}}_9{\text{H}}_{18}{\text{O}}_4}\times 100=\frac{4\times 16.00\text{g/mol}}{\mathrm{180g/mol}}\times 100=\frac{}{}\\ \%\text{O}& \mathrm{35.55\%}\text{O}\end{array}$

QUESTION 14
 Nicotine, an alkaloid in the nightshade family of plants that is mainly responsible for the addictive nature of cigarettes, contains 74.02% C, 8.710% H, and 17.27% N. If 40.57 g of nicotine contains 0.2500 mol nicotine, what is the molecular formula?

Solution
Assuming , a 100-g sample of nicotine yields the following molar amounts of its elements:

$\begin{array}{cc}\left(74.02\text{\%C}\right)=\frac{74.02}{12}& \mathrm{=\; 6.168\; mol}\\ \left(8.710\text{g H}\right)=\frac{8.710}{\mathrm{1.g}}& \mathrm{=\; 8.624}\text{mol}\\ (17.27\text{g N})\; =17.27/14\frac{}{}& \mathrm{=\; 1.233}\text{mol}\end{array}$

Next, we calculate the molar ratios of these elements relative to the least abundant element, N.

$\begin{array}{cc}6.168\text{mol C/}1.233\text{mol =}& 5\\ 8.624\text{mol H/}1.233\text{mol}& \mathrm{=\; 7}\\ 1.233\text{mol N/}1.233\text{mol N}& \mathrm{=\; 1}\end{array}$

We calculate the molar mass for nicotine from the given mass and molar amount of compound:

$\frac{40.57\text{g nicotine}}{0.2500\text{mol nicotine}}=\frac{162.3\text{g}}{\text{mol}}$

Comparing the molar mass and empirical formula mass indicates that each nicotine molecule contains two formula units:

$\frac{162.3}{81\frac{}{}}=2$

Thus, we can derive the molecular formula for nicotine from the empirical formula by multiplying each subscript by two:

$({\text{C}}_5{\text{H}}_7\text{N}{)}_2={\text{C}}_{10}{\text{H}}_{14}{\text{N}}_2$

ASSIGNMENT :
SOLVE NCERT QUESTIONS RELATED TO CALCULATION  CONCENTRATION OF CONCENTRATION OF SOLUTION.

THATS ALL FOR THE DAY.
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