You visited us 5 times! Enjoying our articles? Unlock Full Access!

Elevation in Boiling Point

117 g of NaCl...

Question

$117 g$ of $N a C l$ is added to $222 g$ of water in a container. Calculate the temperature at which the water boils at 1 atm?
Given:
$Ebullioscopic constant for water (K_{b}) = 0.52 K k g {mol}^{- 1}$
$Molecular mass of NaCl = 58.5 g m o l^{- 1}$
and boiling point of pure water is $100^{\circ} C$

{98.3}^{\circ} C

No worries! We‘ve got your back. Try BYJU‘S free classes today!

{102.8}^{\circ} C

No worries! We‘ve got your back. Try BYJU‘S free classes today!

{104.7}^{\circ} C

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

{101.5}^{\circ} C

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C ${104.7}^{\circ} C$
$Weight of solvent w_{1} = 222 g$
$Weight of solute w_{2} = 117 g$
$M_{2} = Molecular mass of NaCl = 58.5 g m o l^{- 1}$
$K_{b} = 0.53 K k g {mol}^{- 1}$

Now, addition of a non-volatile solute causes elevation in boiling point i.e. $Δ T_{b}$
$Δ T_{b} = \frac{(k_{b} \times 1000 \times w_{2})}{(M_{2} \times w_{1})}$
On substituting,
$Δ T_{b} = \frac{(0.52 \times 1000 \times 117)}{(58.5 \times 222)} Δ T_{b} = {4.7}^{\circ} C T_{b} - 100^{\circ} C = {4.7}^{\circ} C$
$T_{b} = (100 + 4.7)^{\circ} C = {104.7}^{\circ} C$ .

Suggest Corrections

Similar questions

Q. The boiling point of pure water is

100^{\circ}

C. Calculate the boiling point of an aqueous solution containing

0.6 g

of urea (molar mass =

60 {g mol}^{- 1}

) in

100 g

of water.

K_{b}

for water is

0.52 {K kg mol}^{- 1}

Q. At higher altitude, boiling point of water is

95^{o} C

. The mass in gram of

N a C l

(

α = 0.8

) added to

100 g

of water

(K_{b} = 0.52 {K kg mol}^{- 1})

in order to raise the boiling point to

100^{o} C

is:

(M.W of N a C l = 58.5 u)

18 g

of glucose,

C_{6} H_{12} O_{6}

(Molar mass = 180 g

m o l^{- 1})

is dissolved in

1 k g

of water in a saucepan. At what temperature will this solution boil?
(

K_{b}

for water =

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

, boiling points of pure water =

373.15 K

)

Boiling point of water at 750 mm Hg is 99.63°C. How much sucrose is to be added to 500 g of water such that it boils at 100°C. Molal elevation constant for water is 0.52 K kg mol⁻¹.

10 g

of a non-electrolyte solute

(M = 60 g/mol)

is added to

250 g

of water. The boiling point of the solution is (

K_{b}

for water is

0.52 {K kg mol}^{- 1}

and boiling point of water

= 373.15 K

)
[0.77 Mark]

Join BYJU'S Learning Program

Explore more

Elevation in Boiling Point

Standard XII Chemistry

Join BYJU'S Learning Program

117 g of NaCl is added to 222 g of water in a container. Calculate the temperature at which the water boils at 1 atm? Given: Ebullioscopic constant for water(Kb)=0.52 K kg mol−1 Molecular mass of NaCl=58.5 g mol−1 and boiling point of pure water is 100∘C

$117 g$ of $N a C l$ is added to $222 g$ of water in a container. Calculate the temperature at which the water boils at 1 atm?
Given:
$Ebullioscopic constant for water (K_{b}) = 0.52 K k g {mol}^{- 1}$
$Molecular mass of NaCl = 58.5 g m o l^{- 1}$
and boiling point of pure water is $100^{\circ} C$