Question

h of the following elements is oxidized and which one is reduced Oxidise- Reduced- 2Mg Oxidise Reduced 4- Complete the table: PH POH OH H+1 1.S 11.6 1023 5. Find the Molarity of 9.8 gram H2SO, in 500 mL of water. 6- How do we prepare a 40 mL solution from 10 mL of 2.5M solution? 7- 4.9 gram of H2SO4 has mixed with water to give a Molarity of 0.2M. How much water is added? A 7 gram of N2 gas is in a 4 liter container with a pressure of 0.4 ATM. Find the Temperature of this ideal gas. 8-

Answer 1

question 3

understanding this qs in a simpler way

oxidation means loss of electron , so if a species looses electron it will gain +ve charge therefore will have increase in oxidation number

wherease reduction means gaining electron , which is accompanied by increase in negative charge hence there will be decrease in oxidation number

The species that will be oxidised is the one which will have increase in its oxidation number and the species which undergoes loss in its oxidation number will be the one which is reduced.

P₄ + 6H₂$\rightarrow$ 4 PH₃

oxidation number of a substance in its elemental state is always zero

so, here oxidation number of ( P₄ ) phosphorous in elemental state is zero

similarly oxidation number of hydrogen (H₂ ) is zero

but in PH₃ , phosphorous shows -3 oxidation number and Hydrogen shows +1/3 oxidation number

So, phosphorous is going from zero oxidation number to -3 that is there is reduction in oxidation number hence it is reduced and hydrogen is going from zero oxidation state to +1/3 this means there is increase in oxidation number of hydrogen and hence it is oxidised .

similarly considering second reaction

2 Mg + O₂$\rightarrow$ 2MgO

here Mg and oxygen are present in their elemental state therefore have zero oxidation number .

in MgO oxidation number of Mg is +2 and of oxygen is -2

therefore Mg is oxidised as there is increase in its oxidation number , and oxygen is reduced as it undergoes decrease in oxidation number.

QUESTION 4

there is very important relation btw PH and POH that is

PH +POH =14

and PH = -log[H⁺]

POH = -log [OH^-]

using these formula to solve the problem

(a)

given PH=1.5

now PH = -log[H⁺]

so, 1.5 = -log[H⁺]

[H⁺] = 10^-1.5 = 0.031622 M

PH + POH = 14

POH = 14 -1.5 = 12.5

POH = -log [OH^-]

[OH^-] = 10^-12.5 = 0.316 $\times$10^-12 M

(b)

given POH =11.6

POH = -log [OH^-]

[OH^-] = 10^-11.6 = 0.2511 $\times$ 10^-11 M

PH+POH =14

PH =14 - 11.6 = 2.4

PH = -log[H⁺]

[H⁺] = 10^-2.4 = 0.003981 M

(c)( The data in image is not clear whether it is 10^-2.5 or 10^-25 )

QUESTION 5

Molarity = number of moles of solute / volume of solution in litres

given

mass of solute = 9.8 g

molar mass of sulphuric acid = 98 g /mol

so , moles of sulphuric acid = 9.8g /98g/mol = 0.1 mol

volume of solution = 500 ml = 0.5 L

molarity = 0.1 mol / 0.5L = 0.2 Molar

QUESTION 6

to prepare 40 ml of solution from 10 ml 2.5 M solution , we simply need to dilute the solution by adding 30 ml of water

QUESTION 7

we are given

mass of solute = 4.9 g

we know molar mass of sulphuric acid = 98 g /mol

so, we can calculate moles of given solute = 4.9g/98g/mol = 0.05 mol

Molarity is given = 0.2 mol/L

volume of solution = ? (we need to calculate)

molarity = number of moles of solute /volume of solution in litres

substituting given values

0.2 mol/ L = 0.05 mol / Volume of solution

Volume of solution = 0.05 mol /0.2 mol/L = 0.25 L

Hence the volume of water added is 0.25 L

QUESTION 8

given 7 gram of N₂ gas in 4 L container with pressure 0.4 atm . we need to calculate temperature

molar mass of N₂ = 28 g/mol

then 7 gram N₂ means = 7 g /28g/mol = 0.25 mol

using ideal gas equation

PV =n RT

p = pressure = 0.4 atm

V = volume = 4 L

n = number of moles of gas = 0.25 mol

R = gas constant (0.0821 L atm mol^-1 K^-1)

T = temperature =? to be calculated

IDEAL GAS EQUATION IS

PV =nRT

susbstituting given values

0.4 atm $\times$ 4L = 0.25 mol $\times$0.0821 L atm mol^-1 K^-1  $\times$ T

T = 77.9 K