The molality of an aqueous dilute solution containing non-volatile solute is 0-1 m- What is the boiling temperature in -C of solution- Boiling point elevation constant- Kb - 0-52 kg mol-1K- boiling temperature of water - 100-C-

The correct option is D 100.052 Given: m=0.1m K_b=0.52 Kg/mol Now, Δ T_b=K_b× m =0.52× 0.1 =0.052^oC Δ T_b=T_b T_b^o ⇒ 0. ...

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Byju's Answer

Standard XII

Chemistry

Molality

The molality ...

Question

The molality of an aqueous dilute solution containing non-volatile solute is $0.1 m$ . What is the boiling temperature (in $^{\circ} C$ ) of solution? (Boiling point elevation constant, $K_{b} = 0.52 k g m o l^{- 1} K$ ; boiling temperature of water $= 100^{\circ} C$ ).

100.0052

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100.052

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100.0

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100.52

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Solution

The correct option is D $100.052$
Given:
$m = 0.1 m$
$K_{b} = 0.52 K g / m o l$
Now,
$Δ T_{b} = K_{b} \times m$
$= 0.52 \times 0.1$
$= {0.052}^{o} C$

$Δ T_{b} = T_{b} - T_{b}^{o}$
$\Rightarrow 0.052 = T_{b} - 100$
$\Rightarrow T_{b} = {100.052}^{o} C$ .

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Similar questions

Q. The molality of an aqueous dilute solution containg non-volatile solute is

0.1

m. What is the boiling temperature ( in

^{o} C

) of solution? (Boiling point elevation constant,

K_{b} = 0.52 k g m o l^{- 1} K

; boiling temperature of water

= 100^{o} C

Q. On mixing

3 g

of non-volatile solute in

200 m L

of water, its boiling point

(100^{\circ} C)

becomes

{100.52}^{\circ} C

. If

K_{b}

for water is

0.6 K / m

then molecular wt. of solute is:

A B_{2}

10 %

dissociated in water to

A^{2 +}

and

B^{-}

. The boiling point of a

10.0

molal aqueous solution of

A B_{2}

^{\circ} C .

(Round off to the nearest Integer).
[Given: Molal elevation constant of water

K_{b} = 0.5 {K kg mol}^{- 1}

, boiling point of pure water

= 100^{\circ} C

]

Q. The boiling point of an aqueous solution of a non-volatile solute is

{100.15}^{\circ} C

. What is the freezing point of an aqueous solution obtained by diluting the above solution with an equal volume of water? The values of

k_{b}

and

k_{f}

for water are

0.512

and

1.86 K {molality}^{- 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 atm pressure is $2^{\circ} C$ . Assuming concentration of solute is much lower than the concentration of solvent, the vapour pressure (mm of Hg) of the solution is (take $K_{b}$ =0.76K kg $m o l^{- 1}$ ).

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The molality of an aqueous dilute solution containing non-volatile solute is 0.1 m. What is the boiling temperature (in ∘C) of solution? (Boiling point elevation constant, Kb=0.52 kg mol−1K; boiling temperature of water =100∘C).

The correct option is D 100.052Given:m=0.1mKb=0.52Kg/molNow,ΔTb=Kb×m=0.52×0.1=0.052oCΔTb=Tb−Tob⇒0.052=Tb−100⇒Tb=100.052oC .

The molality of an aqueous dilute solution containing non-volatile solute is $0.1 m$ . What is the boiling temperature (in $^{\circ} C$ ) of solution? (Boiling point elevation constant, $K_{b} = 0.52 k g m o l^{- 1} K$ ; boiling temperature of water $= 100^{\circ} C$ ).

The correct option is D $100.052$
Given:
$m = 0.1 m$
$K_{b} = 0.52 K g / m o l$
Now,
$Δ T_{b} = K_{b} \times m$
$= 0.52 \times 0.1$
$= {0.052}^{o} C$

$Δ T_{b} = T_{b} - T_{b}^{o}$
$\Rightarrow 0.052 = T_{b} - 100$
$\Rightarrow T_{b} = {100.052}^{o} C$ .