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Chemistry Equilibrium

pH value of 0.001 M NaOH solution is___________.

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Payal Gupta

Contributor-Level 10

0.001 M NaOH solution has [OH^-] = 0.001 M = 10^-³ M

Using ; pOH = $- l o g_{10} [O H^{-}]$

= $- l o g_{10} 1 0^{- 3}$

pOH = 3

pH = 14 – pOH

pH = 14 – 3

pH = 11

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11 months ago

Chemistry Solutions

A solution containing 2.5 * 10^-3 kg of solute dissolved in 75 * 10^-3 kg of water boils at 373.535 K. The molar mass of the solute is_________g mol^-1. (nearest integer)

[Given K_b(H₂O) = 0.52 K kg mol^-1 and boiling point of water = 373.15 K)

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Payal Gupta

Contributor-Level 10

Mass of solute ; w_B = 2.5 * 10^-³ kg

Mass of solvent, $w_{A} = 75 * 1 0^{- 3} k g$

Boiling point of solution ; $T_{b} = 373.535 K$

Boiling point of water ; $T_{b}^{0} = 373.15 K$

So, elevation in boiling point, $Δ T_{b} = T_{b} - T_{b}^{0}$

$Δ T_{b} = 0.385 K$

Using ; $Δ T_{b} = K_{b} * \frac{w_{B}}{M_{B} * w_{A}} * 1000$

$0.385 = 0.52 * \frac{2.5 * 1 0^{- 3}}{M_{B} * 75 * 1 0^{- 3}} * 1000$

Molar mass of solute ; M_B = 45 g/mol

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Chemistry Chemistry NCERT Exemplar Solutions Class 12th Chapter Six

When 5 moles of He gas expand isothermally and reversibly at 300K from 10 litre to 20 litre, the magnitude of the maximum work obtained is___________J. (nearest integer)

[Given: R=8.3 JK^-1mol^-1 and log2 = 0.3010]

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Payal Gupta

Contributor-Level 10

Number of moles, n = 5 mol

Temperature, T= 300 K

Initial volume, V₁ = 10L

Final volume, V₂ = 20 L

Using;

Work done; w = -2.303 nRT log₁₀ $\frac{V_{2}}{V_{1}}$

$= - 2.303 * 5 * 8 * 300 l o g_{10} \frac{20}{10} J$

= -8630 J

So, magnitude of work done is 8630 J.

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11 months ago

Chemistry Chemistry NCERT Exemplar Solutions Class 12th Chapter Ten

BeO reacts with HF in presence of ammonia to give [A] which on thermal decomposition produces [B] and ammonium fluoride. Oxidation state of Be in [A] is____________.

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Payal Gupta

Contributor-Level 10

$B e O + 2 N H_{3} + 4 H F \to {(N H_{4})}_{2} [B e F_{4}]$

${(N H_{4})}_{2} [B e F_{4}] \to_{}^{Δ} B e F_{2} + 2 N H_{4} F$

here, oxidation state of be in A is +2.

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11 months ago

Chemistry Structure of Atom

Consider the following set of quantum numbers.

	n	$l$	$m_{l}$
A.	3	3	-3
B.	3	2	-2
C.	2	1	+1
D.	2	2	+2

The number of correct sets of quantum numbers is____________.

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Payal Gupta

Contributor-Level 10

n = 1, 2, 3 …….; $l = 0$ ……. to n – 1, $m_{l} = l . . . . . . . 0 . . . . . . + l$

A. n = 3 $l = 3$ $m_{l} = - 3$ is incorrect as $l$ can not be equal to n.

B. n = 3 $l = 2$ $m_{l}$ = -2 is correct set.

C. n = 2 $l = 1$ $m_{l}$ = +2 is incorrect set as $l$ can not be equal to n.

So; correct set of quantum numbers is B and C.

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11 months ago

Chemistry Some Basic Concepts of Chemistry

116 g of a substance upon dissociation reaction yields 7.5 g of hydrogen 60 g of oxygen and 48.5 g of carbon. Given that the atomic masses of H, O and C are 1, 16 and 12, respectively. The data agrees with how many formulae of the following?

(a) CH₃COOH (b) HCHO (c) CH₃OOCH₃ &n

...more

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Payal Gupta

Contributor-Level 10

Determining empirical formula of the given compound.

$% o f C = \frac{48.5}{116} * 100 % = 41.8 %$

$% o f H = \frac{7.5}{116} * 100 % = 6.5 %$

$% o f O = \frac{60}{116} * 100 % = 51.7 %$

Mass Moles Simplest ratio

C 41.8g $\frac{41.8}{12} = 3.48$ $\frac{3.48}{3.23} \approx 1$ &n

...more

Determining empirical formula of the given compound.

$% o f C = \frac{48.5}{116} * 100 % = 41.8 %$

$% o f H = \frac{7.5}{116} * 100 % = 6.5 %$

$% o f O = \frac{60}{116} * 100 % = 51.7 %$

Mass Moles Simplest ratio

C 41.8g $\frac{41.8}{12} = 3.48$ $\frac{3.48}{3.23} \approx 1$

H 6.5 g $\frac{6.5}{1} = 6.5$ $\frac{6.5}{3.23} \approx 2$

O 51.7 g $\frac{51.7}{16} = 3.23$ $\frac{3.23}{3.23} = 1$

Ratio of moles of C, H and O is 1 : 2 : 1

A. CH₃COOH ⇒ C : H : O = 2 : 4 : 2 = 1 : 2 : 1

B. HCHO ⇒ C : H : O = 1 : 2 : 1

C. CH₃OOCH₃ ⇒ C : H : O = 2 : 6 : 2 = 1 : 3 : 1

D. CH₃CHO ⇒ C : H : O = 2 : 4 : 1

So, CH₃COOH and HCHO formulae agree with the given data

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11 months ago

Chemistry Chemistry in Everyday Life

Match List I with List II

List I		List II
(Anion)		(gas evolved on reaction with dil H₂SO₄)
A.	CO₃^2-	I.	Colourless gas which turns lead acetate paper black
B.	S²^-	II.	Colourless gas which turns acidified potassium dichromate solution green
C.	SO₃²-	III.	Brown fumes which turns acidified Kl solution containing starch blue.
D.	NO₂^-	IV.	Colourless gas evolved with brisk effervescence, which turns lime water milky

Choose the correct answer from the option given below:

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Payal Gupta

Contributor-Level 10

A. $\underset{S a l t}{N a_{2} C O_{3}} + \underset{d i l}{H_{2} S O_{4}} \to N a S O_{4} + H_{2} O + C O_{2} ↑$

CO₂ is colourless gas with brisk effervescence and turns lime water milky

$C O_{2} + C a {(O H)}_{2} \to C a C O_{3} ↓ + H_{2} O$

B. $\underset{S a l t}{N a_{2} S} + \underset{d i l}{H_{2} S O_{4}} \to N a_{2} S O_{4} + H_{2} S ↑$

H₂S + (CH₃COO)₂Pb → $\underset{B l a c k}{P b S} + 2 C H_{3} C O O H$

C. $\underset{S a l t}{N a_{2} S O_{3}} + \underset{d i l}{H_{2} S O_{4}} \to N a_{2} S O_{4} + H_{2} O + S O_{3} ↑$

SO₃ is colourless gas which turns acidified potassium dichromate solution green as

$S O_{3} + K M n O_{4} + H^{+} \to \underset{G r e e n}{M n S O_{4}} + S O_{4}^{2 -} + H_{2} O$

D. $\underset{S a l t}{2 N a N O_{2}} + \underset{d i l}{H_{2} S O_{4}} \to N a_{2} S O_{4} + 2 H N O_{2}$ &nbs

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A. $\underset{S a l t}{N a_{2} C O_{3}} + \underset{d i l}{H_{2} S O_{4}} \to N a S O_{4} + H_{2} O + C O_{2} ↑$

CO₂ is colourless gas with brisk effervescence and turns lime water milky

$C O_{2} + C a {(O H)}_{2} \to C a C O_{3} ↓ + H_{2} O$

B. $\underset{S a l t}{N a_{2} S} + \underset{d i l}{H_{2} S O_{4}} \to N a_{2} S O_{4} + H_{2} S ↑$

H₂S + (CH₃COO)₂Pb → $\underset{B l a c k}{P b S} + 2 C H_{3} C O O H$

C. $\underset{S a l t}{N a_{2} S O_{3}} + \underset{d i l}{H_{2} S O_{4}} \to N a_{2} S O_{4} + H_{2} O + S O_{3} ↑$

SO₃ is colourless gas which turns acidified potassium dichromate solution green as

$S O_{3} + K M n O_{4} + H^{+} \to \underset{G r e e n}{M n S O_{4}} + S O_{4}^{2 -} + H_{2} O$

D. $\underset{S a l t}{2 N a N O_{2}} + \underset{d i l}{H_{2} S O_{4}} \to N a_{2} S O_{4} + 2 H N O_{2}$

$3 H N O_{2} \to N H O_{3} + 2 N O + H_{2} O$

$2 N O + O_{2} \to \underset{B r o w n g a s}{2 N O_{2}}$

$2 H N O_{2} + 2 K l + 2 C H_{3} C O O H \to 2 C H_{3} C O O K + 2 H_{2} O + 2 N O + I_{2}$

I₂ + starch → Blue colour complex.

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11 months ago

Chemistry Chemistry in Everyday Life

Match List I with List II

List I List II

A. Antipyretic I. Reduces pain

B. Analgesic II. Reduces stress

C. Tranquilizer &nbs

...more

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Payal Gupta

Contributor-Level 10

A. Antipyretic reduces fever.

B. Analgesic reduces pain.

C. Tranquilizer reduces stress.

D. Antacid reduces acidity of stomach.

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Chemistry Biomolecules

Given below are two statements.

Statements I: Maltose has two a-D-glucose units linked at C₁ and C₄ and is a reducing sugar:

Statements II: Maltose has two manosaccharides: a-D-glucose and b-D-glucose linked at C₁ and C₆ and its is a non-reducing sugar.

In the light of the above statements, choose the correct answer from the options given below.

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Payal Gupta

Contributor-Level 10

$C_{12} H_{22} O_{11} + H_{2} O \to α - D - g l u c o s e + α - D - g l u c o s e$

In maltose, two units of a-D-glucose are linked by glycosidic linkage at C₁ and C₄ as shown.

a - D – glucose a - D - glucose

Maltose has hemiacetal link so it is reducing sugar.

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11 months ago

Chemistry Polymers

Which is true about Buna-N

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Payal Gupta

Contributor-Level 10

Buna-N is obtained by copolymerization of 1, 3-butadiene and acrylonitrile,

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