Boiling point elevation is given as:
ΔTb=iKbm
Where,
ΔTb=elevation in the boiling point
that is given by expression:
ΔTb=Tb (solution) - Tb (pure solvent)
Here Tb (pure solvent)=118.1 °C
i for CaCO3= 2
Kb=2.93 °C/m
m=Molality of CaCO₃:
Molality of CaCO₃=Number of moles of CaCO₃/ Mass of solvent (Kg)
=(Given Mass of CaCO3/Molar mass of CaCO₃)/ Mass of solvent (Kg)
=(100.0÷100 g/mol)/0.4
= 2.5 m
So now putting value of m, i and Kb in the boiling point elevation equation we get:
ΔTb=iKbm
=2×2.93×2.5
=14.65 °C
boiling point of a solution can be calculated:
ΔTb=Tb (solution) - Tb (pure solvent)
14.65=Tb (solution)-118.1
Tb (solution)=118.1+14.65
=132.75
Answer:
17.5609g
Explanation:
According to the question, a sample of mass 6.814 grams is added to another sample weighing 0.08753 grams. That is weight of sample 1 + weight of sample 2;
6.814 + 0.08753 = 6.90153grams
Next, the subsequent mixture is then divided into exactly 3 equal parts i.e. 6.90153grams divided by 3
= 6.90153/3
= 2.30051grams.
One of the equal parts is 2.30051grams, which is then multiplied by 7.6335 times I.e. 2.30051 × 7.6335 = 17.5609grams
Therefore, the final mass is 17.5609grams
Answer: The pressure will be 18.05 atm.
Explanation: Expression for ideal gas equation is :

where,
P = Pressure of the gas = ? atm
V = Volume of the gas = 0.333L
n = Number of moles of gas = 0.250 moles
R = Universal gas constant = 
T = temperature of the gas = 20°C = (273 + 20)K = 293K
Putting values in above equation, we get:

P = 18.05 atm
The density of a material is the mass of the material per unit volume. Here the weight of the same metal is 44.40g, 40.58g and 38.35g having volume 4.8 mL, 4.7 mL and 4.2 mL respectively. Thus the density of the metal as per the given data are,
= 9.25g/mL,
= 8.634g/mL and
= 9.130g/mL respectively.
The equation of the standard deviation is √{∑(x -
)÷N}
Now the mean of the density is {(9.25 + 8.634 + 9.130)/3} = 9.004 g/mL.
The difference of the density of the 1st metal sample (9.25-9.004) = 0.246 g/mL. Squaring the value = 0.060.
The difference of the density of the 2nd metal sample (9.004-8.634) =0.37 g/mL. Squaring the value = 0.136.
The difference of the density of the 3rd metal sample (9.130-9.004) = 0.126 g/mL. Squaring the value 0.015.
The total value of the squared digits = (0.060 + 0.136 + 0.015) = 0.211. By dividing the digit by 3 we get, 0.070. The standard deviation will be
. Thus the standard deviation of the density value is 0.265g/mL.