18.
A solution of sucrose (molar mass=342) has been prepared by dissolving
68.4
g of sucrose in 1 kg of water. Ke for water is 1.86 k kg mol-!
&
vapour pressure of water at 298 k is 0.024 atm. Calculate the :
(i) Vapour pressure of solution at 298 k &
(ii) The freezing point of solution.
(2006 Compartment)
Answers
Answer:
Answer:
Push or pull of an object is considered a force. Push and pull come from the objects interacting with one another. Terms like stretch and squeeze can also be used to denote force.
In Physics, force is defined as:
The push or pull on an object with mass that causes it to change its velocity.
Force is an external agent capable of changing the state of rest or motion of a particular body. It has a magnitude and a direction. The direction towards which the force is applied is known as the direction of the force and the application of force is the point where force is applied.
The Force can be measured using a spring balance. The SI unit of force is Newton(N).
Common symbols: F→, F
SI unit: Newton
In SI base units: kg·m/s2
Other units: dyne, poundal, pound-force, kip, kilo pond
Derivations from other quantities: F = m a
Dimension: LMT-2Answer: \dfrac{1316}{3}
3
1316
Step-by-step explanation:
To find: 1 \div 1 \times 2\times 3+1 \div2 \times3 \times4 + 1 \div3\times 4 \times 5+ 14\times 5 \times 61÷1×2×3+1÷2×3×4+1÷3×4×5+14×5×6
\begin{gathered}1 \div 1 \times 2\times 3+1 \div2 \times3 \times4 + 1 \div3\times 4 \times 5+ 14\times 5 \times 6\\\\= \dfrac{1}{1} \times 6+\dfrac{1}{2} \times 12+ \dfrac{1}{3} \times 20 + 420\\\\=6+6+\dfrac{20}{3} +420\\\\=432+\dfrac{20}{3}\\\\=\dfrac{1296+20}{3} \\\\=\dfrac{1316}{3}\end{gathered}
1÷1×2×3+1÷2×3×4+1÷3×4×5+14×5×6
=
1
1
×6+
2
1
×12+
3
1
×20+420
=6+6+
3
20
+420
=432+
3
20
=
3
1296+20
=
3
1316
Hence the answer is \Draco{1316}{3}
3
1316
Answer:
ANSWER
DepressioninFreezingpointΔT
f
=
M
1
×W
2
W
1
×K
f
×1000
whereW
1
=Weight of Solute
W
2
=Weight of solvent
M
1
=Molar mass of solute
K
f
=Freezing point deprssion constant
Now,ΔT
f
=
342×1000
1.86×68.5×1000
=0.372C
Now,ΔT
f
=T
o
−T
f
So,T
f
=0−0.372=−0.372C.
(Freezing point of purewater=0
0
C.)
Hence, the correct option is A