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Standard XII

Chemistry

Elevation in Boiling Point

1 g of non-vo...

Question

$1 g$ of non-voltaile non-electrolyte solute is dissolved in $100 g$ of two different solvents $A$ and $B$ whose ebullioscopic constants are in the ratio of $1 : 5$ . The ratio of the elevation in their boiling points, $\frac{△ T_{b} (A)}{△ T_{b} (B)}$ , is:

5 : 1

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10 : 1

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1 : 5

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1 : 0.2

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Solution

The correct option is C $1 : 5$
We know,
$△ T_{b} = K_{b} \times m$
$∴ \frac{△ T_{b (A)}}{△ T_{b (B)}} = \frac{K_{b (A)}}{K_{b (B)}}$ as $m_{A} = m_{B}$
$∴ \frac{△ T_{b (A)}}{△ T_{b (B)}} = \frac{1}{5}$ .

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Similar questions

Q. 10 gram of solute with molecular mass 100 gram

m o l^{- 1}

is dissolved in a 100-gram solvent to show

{0.3}^{o} C

elevation in boiling point. The value of molal ebullioscopic constant will be:

Q. If

1 g

of solute (molar mass

= 50 g

{m o l}^{- 1}

) is dissolved in

50 g

of solvent and the elevation in boiling point is

1 K

. The molar boiling constant of the solvent is?

Q. If the elevation in boiling point of a solution of non-volatile, non-electrolytic and non-associating solute in a solvent

(K_{b} = x K k g m o l^{- 1})

y K

, then the depression in freezing point of solution of same concentration would be:

[K_{f}

of the solvent

= z K k g m o l^{- 1}]

For a dilute solution containing 2.5 g of a non-volatile non-electrolyte solute in $100 g$ of water, the elevation in boiling point at $1 a t m$ pressure is $2 ℃$ . Assuming that the concentration of solute is much lower than the concentration of solvent, the vapour pressure $(in mm of Hg)$ of the solution is:

[Take : $K_{b} = 0.76 K k g m o l^{- 1}$ ]

Q. Assertion :Molal elevation constant depends on the nature of the solvent.
Reason: Molal elevation constant is the elevation in boiling point when

1 m o l e

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1 k g

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1 g of non-voltaile non-electrolyte solute is dissolved in 100 g of two different solvents A and B whose ebullioscopic constants are in the ratio of 1:5. The ratio of the elevation in their boiling points, △Tb(A)△Tb(B), is:

The correct option is C 1:5We know,△Tb=Kb×m∴△Tb(A)△Tb(B)=Kb(A)Kb(B) as mA=mB∴△Tb(A)△Tb(B)=15.

$1 g$ of non-voltaile non-electrolyte solute is dissolved in $100 g$ of two different solvents $A$ and $B$ whose ebullioscopic constants are in the ratio of $1 : 5$ . The ratio of the elevation in their boiling points, $\frac{△ T_{b} (A)}{△ T_{b} (B)}$ , is:

The correct option is C $1 : 5$
We know,
$△ T_{b} = K_{b} \times m$
$∴ \frac{△ T_{b (A)}}{△ T_{b (B)}} = \frac{K_{b (A)}}{K_{b (B)}}$ as $m_{A} = m_{B}$
$∴ \frac{△ T_{b (A)}}{△ T_{b (B)}} = \frac{1}{5}$ .