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Standard VIII

Chemistry

Solvent and Solute

what is the m...

Question

what is the ma in difference between solute and solvent ?

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Solution

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Similar questions

Q. What are the differences between a solute/solvent and solution/suspension?

Q. The vapour pressure of a solution of a non-volatile electrolytes (A) in a solvent (B) is

95

% of the vapoure pressure of the solvent at the same temperature. If

M_{B} = 0.3 M_{A}

, where

M_{B}

and

M_{A}

are molar masses of

B

and

A

respectively, the mass ratio of the solvent and solute are:

Molality: It is defined as the number of moles of the solute present in 1 kg of the solvent. It is denoted by 'm'
Molality (m) =

\frac{N u m b e r o f m o l e s o f s o l u t e}{N u m b e r o f k i l o g r a m s o f t h e s o l v e n t}

Let

w_{A}

grams of the solute of molecular mass

m_{A}

be present in

w_{B}

grams of the solvent, then
Molality (m) =

\frac{w_{A}}{m_{A} \times w_{B}} \times 1000

Relation between mole fraction and Molality :

X_{A} = \frac{n}{N + n} a n d X_{B} = \frac{N}{N + n}

\frac{X_{A}}{X_{B}} = \frac{n}{N} = \frac{M o l e s o f s o l u t e}{M o l e s o f s o l v e n t} = \frac{W_{A} \times m_{B}}{w_{B} \times m_{A}}

\frac{X_{A} \times 1000}{X_{B} \times m_{B}} = \frac{w_{A} \times 1000}{W_{B} \times m_{A}} = m

, or,

\frac{X_{A} \times 1000}{(1 - X_{A}) m_{B}} = m

If the ratio of the mole fraction of a solute is changed from

\frac{1}{3}

\frac{1}{2}

in the

800

g of solvent then, the ratio of molality will be :

Q. The vapour pressure of a solution of a non - volatile electrolyte (A) in a solvent (B) is 95% of the vapour pressure of the solvent at the same temperature. If

M_{B} = 0.3 M_{A}

where

M_{B}

and

M_{A}

are molecular weights of (B) and (A) respectively, the weight ratio of the solvent and solute are:

Question 33
In isotonic solutions ........... .
(a) solute and solvent both are same
(b) osmotic pressure is same
(c) solute and solvent may or may not be same
(d) solute is always same solvent may be different

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