CameraIcon

SearchIcon

MyQuestionIcon

4

You visited us 4 times! Enjoying our articles? Unlock Full Access!

Elevation in Boiling Point

Which solutio...

Question

Which solution would probably have the highest boiling point?

A

0.100

m KOH

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

B

0.100

m

N a_{2} S O_{4}

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

C

0.100

m

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

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

D

0.200

m

C a C l_{2}

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

E

0.200

m

C H_{3} C H_{2} O H

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

Open in App

Solution

The correct option is D $0.200$ m $C a C l_{2}$
$0.200 m C a C l_{2}$ solution would probably have the highest boiling point.
The elevation in the boiling point is directly proportional to the product $i \times m$ where 'i' is vant Hoff's factor and m is the molality of solution. This product is maximum for $0.200 m C a C l_{2}$ solution. Hence, it gives maximum boiling point.
Note: 1 molecule of $C a C l_{2}$ dissociates in aqueous solution to give 3 ions.
Hence, $i \times m$ for $0.200 m C a C l_{2} = 0.200 \times 3 = 0.600$

Suggest Corrections

thumbs-up

0

Similar questions

Q. Determine the rate law for

A + B \to C

based on the following information.

$[A]$	$[B]$	Initial Rate (M/s)
$0.200$	$0.100$	$4.0 \times 10^{- 3}$
$0.200$	$0.200$	$4.0 \times 10^{- 3}$
$0.400$	$0.100$	$1.6 \times 10^{- 2}$

40.0 m L

0.100 M H N O_{3}

is titrated with

0.200 M N a O H

.
Calculate the volume required to reach the equivalence point.

What volume of 0.50 M $C a C l_{2}$ solution is needed to prepare 250 mL of solution that has a chloride concentration of 0.100 M?

What is the significant figure of ?:

1. 100

2. 20

3. 0.100

4. 0.200

Q. The table below shows recorded concentration data for the following chemical reaction:

C a C l_{2} (a q) + 2 A g N O_{3} (a q) \to C a (N O_{3})_{2} (a q) + 2 A g C l (s)

Use the data to determine the exponents

x

y

in the rate law:

r a t e = k [C a C l_{2}]^{x} [A g N O_{3}]^{y}

$[C a C l_{2}] M$	$[A g N O_{3}] M$	Rate M/s
$0.050$	$0.050$	$3.33 \times 10^{- 3}$
$0.100$	$0.050$	$6.66 \times 10^{- 3}$
$0.100$	$0.100$	$13.32 \times 10^{- 3}$

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Elevation in Boiling Point

CHEMISTRY

Watch in App

explore_more_icon

Explore more

Elevation in Boiling Point

Standard XII Chemistry

Join BYJU'S Learning Program

CrossIcon