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73.606 °C is the freezing point of the solution made with with 1.31 mol of CHCl3 in 530.0 g of CCl4.
Explanation:
Data given:
number of moles of CHCl3 = 1.31 moles
mass of solvent CHCl3 = 530 grams or 0.53 kg
Kf = 29.8 degrees C/m
freezing point of pure solvent or CCl4 = -22.9 degrees
freezing point = ?
The formula used to calculate the freezing point of the mixture is
ΔT = iKf.m
m= molality
molality =
putting the value in the equation:
molality=
= 2.47 M
Putting the values in freezing point equation
ΔT = 1.31 x 29.8 x 2.47
ΔT = 73.606 degrees
11.48-gram of are needed to produce 6.75 Liters of
gas measured at 1.3 atm pressure and 298 K
The ideal gas law (PV = nRT) relates the macroscopic properties of ideal gases. An ideal gas is a gas in which the particles (a) do not attract or repel one another and (b) take up no space (have no volume).
First, calculate the moles of the gas using the gas law,
PV=nRT, where n is the moles and R is the gas constant. Then divide the given mass by the number of moles to get molar mass.
Given data:
P= 1.3 atm
V= 6.75 Liters
n=?
R=
T=298 K
Putting value in the given equation:
Moles = 0.3588 moles
Now,
Mass= 11.48 gram
Hence, 11.48-gram of are needed to produce 6.75 Liters of
gas measured at 1.3 atm pressure and 298 K
Learn more about the ideal gas here:
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