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212 ml of lead nitrate is required to prepare a dilute solution of 820.7 ml of lead nitrate.
Answer:
Option A.
Explanation:
Similar to Avagadro's law, there is another law termed as dilution law. As the product of volume and normality of the reactant is equal to the product of volume and normality of the product from the Avagadro's law. In dilution law, it will be as product of volume and concentration of the solute of the reactant is equal to the product of volume and concentration of solution.
So, as per the given question C1 = 5.45 M of lead nitrate and V1 has to be found. While C2 is 1.41 M of lead nitrate and V2 is 820.7 ml.
Then,
So nearly 212 ml of lead nitrate is required to prepare a dilute solution of 820.7 ml of lead nitrate.
Answer:
4.8 g/mL is the density of chloroform vapor at 1.00 atm and 298 K.
Explanation:
By ideal gas equation:
Number of moles (n)
can be written as:
where, m = given mass
M = molar mass
where,
which is known as density of the gas
The relation becomes:
.....(1)
We are given:
M = molar mass of chloroform= 119.5 g/mol
R = Gas constant =
T = temperature of the gas =
P = pressure of the gas = 1.00 atm
Putting values in equation 1, we get:
4.8 g/mL is the density of chloroform vapor at 1.00 atm and 298 K.