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# the concentration of sucrose solution which is isotonic with 0.02 m

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### Mohammed

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## 2. The concentreatiln of sucrose solution which is isotonic with 0.02 M solution of Na2SO4 T 30 degree celcius

2. The concentreatiln of sucrose solution which is isotonic with 0.02 M solution of Na2SO4 T 30 degree celcius Byju's Answer Standard VII Chemistry Indicators 2. The concen... Question

2. The concentreatiln of sucrose solution which is isotonic with 0.02 M solution of Na2SO4 T 30 degree celcius

Open in App Solution Suggest Corrections 16 SIMILAR QUESTIONS

Q.

The percentage of urea in an aqueous solution which is isotonic with a 34.2% solution of sucrose is

Q. Which of the following solutions are isotonic with respect to

0.5 M N a C l

solution having the degree of dissociation = 0.8.

Q.

A 0.004 M N a 2 S O 4

aqueous solution is isotonic with

0.01 M glucose at 300 K

. Thus, the degree of dissociation of

N a 2 S O 4 is

Q.

An aqueous solution of sucrose (C12H22O11) containing 34.2 g/L of sucrose has an osmotic pressure of 2.38 atm at 17° C. For an aqueous solution of glucose (C6H12O6) to be isotonic with this solution, it's concentration (in g/L) should be:

Q. A

0.004 M solution of N a 2 S O 4 is isotonic with a 0.010 M

solution of glucose at same temperature. The apparent precentage degree of dissociation of

N a 2 S O 4 is: View More

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## What is molar concentration of Ca(OH)2 if its solution has pH of 12 ?

Click here👆to get an answer to your question ✍️ The concentration of sucrose solution which is isotonic with 0.02 M solution of Na, So, (assume complete dissociation) at 30°C is (1) 0.04 M (2) 0.02 M (3) 0.06M (4) 0.08 M Question ## The concentration of sucrose solution which is isotonic with 0.02 M solution of Na, So, (assume complete dissociation) at 30°C is (1) 0.04 M (2) 0.02 M (3) 0.06M (4) 0.08 M

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Updated on : 2022-09-05

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## The concentration of sucrose solution which is isotonic with 002M solution of Na2SO4 assume complete dissociation at 30C is

Isotonic solution have same osmotic pressure iCRT WhereRandTare constant iVanthoff factor The dissociation ofNa2SO4is as follows Na2SO42NaSO42 Soi3 sucroseNa2SO4 1C3002 C006M So concentration of sucrose solution is006M

Gamma Question Bank for Medical Chemistry>Solutions>Assignment>Q 12

EASY Earn 100

The concentration of sucrose solution which is isotonic with

0.02M solution of Na2SO4

(assume) complete dissociation) at

30°C is 100%

## Important Questions on Solutions

EASY

Gamma Question Bank for Medical Chemistry>Solutions>Assignment>Q 13

The effect of increase in the temperature of an aqueous solution of a substance is

EASY

Gamma Question Bank for Medical Chemistry>Solutions>Assignment>Q 14

When a solute associates in solution, then van't Hoff factor (i) is

EASY

Gamma Question Bank for Medical Chemistry>Solutions>Assignment>Q 15

If degree of dissociation of

Al2SO43 is 25% in a solvent, then

## Important Points to Remember on Solutions

1. Solution:

A homogeneous mixture of two or more non-reacting substances is known as solution, Homogeneity or heterogeneity depends upon particle size and states of matter present in the solution. Every solution is made up of a solvent (present in larger quantity) and one or more solute (present in smaller quantity).

2. Units of Concentration:

(i) Molarity M :

It is the number of moles of solute present per litre of solution.

M=nV=WMwVin litre=WMw1000Vin cc

M×Vin cc=WMw1000 ⇒M×VmL= millimoles

Molarity changes with temperature of the solution. Increase in temperature generally decreases the molarity. It is the most convenient method to express concentration of the solution. On dilution, molarity decreases.

(ii) Molality m : Number of moles n

of solute present per

kg of solvent.

m=nWin kg=WMWin kg=WMWin gsolvent 1000

It is independent of temperature since no volume factor is involved in the equation.

(iii) Mole fraction X :

It is the ratio of number of moles of one component to the total number of moles present in the solution.

For a system having two components

A and B ,

XA=nAnA+nB,  XB=nBnA+nB

XA+XB=1

Mole fraction is also independent of temperature.

(iv) In terms of % % by weight

= Wt. of solute  Wt. of solution ×100

% weight by volume

= Wt.of solute  Vol. of solution ×100

(In case of solid dissolved in a liquid)

% by volume

= Volume of solute  Volume of solution ×100

(In case of liquid dissolved in another liquid)

PPM= No. of parts of solute  No. of parts of solution ×106

%

by weight is independent of temperature while

%

by volume are temperature dependent.

3. Henry’ s Law:

Solubility of a gas at a given temperature in a solvent is directly proportional to its partial pressure, if

P

is the partial pressure of a gas and

Xg

is its mole fraction in solution. Then

P=KHXg , where KH

is Henry's law constant for that gas.

4. Vapour Pressure and Raoult’s law:

The pressure exerted by the vapour at the free surface of liquid (provided system is closed) is known as its vapour pressure. The

V.P.

of a pure liquid is always greater than its solution (in case of non-volatile solute).

(i) Raoult’s Law for a solution having non-volatile solute.

Xsolute =P∘-PsP∘ Xsolute →

Mole fraction of solute in solution

P∘→ V.P. of pure solvent Ps→ V.P. of solution

i.e., relative lowering of vapour pressure is equal to the mole fraction of solute.

(ii) Raoult's Law of miscible liquid-liquid solution

For ideal solution, the partial vapour pressure is directly proportional to their mole fraction at constant temperature. For two components

A and B in liquid solution. ⇒PA=PA∘XA ⇒PB=PB∘XB The total pressure

PTotal=PA+PB=PA∘XA+PB∘XB Most of the solutions show appreciable deviations from ideal behavior known as real or non-ideal solution. In some cases, the deviation is

+ve

while in some cases deviation is

-ve .

5. Ideal and non-ideal Solutions:

The solutions which obey Raoult's law are ideal solutions and those which do not obey Raoult's law form non-ideal solution.

Ideal Solution Non-Ideal Solution

Positive Deviation Negative Deviation

1. Obey Raoult's law Disobey Raoult's law Disobey Raoult's law

pA

2. pA=pA∘XA; pB=pB∘XB pTotal=pA+pB pA>pA∘XA;  pB>pB∘XB pTotal=pA+pB pTotal>PA+PB

pTotal

pTotal=pA+pB

A-B

3. ΔHmix =0 ΔVmix =0 ΔGmix=-ve ΔSmix = +ve ΔHmix =+ve ΔVmix =+ve ΔGmix=-ve ΔSmix = +ve ΔHmix =-ve ΔVmix =-ve ΔGmix=-ve ΔSmix = +ve 4. Interaction A-B=A-A=B-B e.g., Chlorobenzene + Bromobenzene Interaction

and B-B e.g., CH3OH+H2O Interaction A-B>A-A and B-B e.g., CH3COCH3+CHCl3

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Mohammed 1 month ago

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