
What is
(i) the molar mass of lead nitrate, and
(ii) the molar quantity of a \[42.{\text{ }}65{\text{ }}g\] mass of this material?
Answer
470.1k+ views
Hint: We need to know the definition of molar mass and the composition of lead nitrate. The molar quantity of lead nitrate is also to be calculated. With the given molecular formula of a compound and the atomic masses of its constituent elements, we calculate the molar mass of a given compound. The molar mass will also be used to calculate the molar quantity.
Formula to be used:
\[Molar{\text{ }}quantity = {\text{ }}\dfrac{{Given{\text{ }}mass}}{{{\text{ M}}olar{\text{ }}mass}}\]
Complete answer:
(i) The following rules are used to compute a compound's molecular mass:
1. The atomic masses of the elements that make up the specified compound must be known.
2. We multiply the atomic mass of a certain element by the number of instances of that element in the compound (written in the form of subscript of the element in the molecular formula).
3. We add all of the results together to get the molecular mass in grams/mole.
The molecular formula of lead nitrate is $Pb{(N{O_3})_2}$whose molar mass is calculated as $\{ 207.2 + 2 \times 14.01 + 6 \times 15.999\} g \cdot mo{l^{ - 1}}$
= $331.214gmo{l^{ - 1}}$
(ii) The molar quantity of a \[42.{\text{ }}65{\text{ }}g\]mass of lead nitrate is as follows:
\[Molar{\text{ }}quantity = {\text{ }}\dfrac{{Given{\text{ }}mass}}{{{\text{ M}}olar{\text{ }}mass}}\]
The given mass of the material is \[42.{\text{ }}65{\text{ }}g\]and the molar mass is $331.214gmo{l^{ - 1}}$. Putting these values, we get
\[Molar{\text{ }}quantity = {\text{ }}\dfrac{{42.65}}{{{\text{ 331}}{\text{.214}}}}\]
\[ = 0.128mol\]
Note:
It's worth mentioning that a compound's molar mass is known by a variety of common names. Among other names, the density of one mole of material in grams per mole is referred to as molecular mass, molecular weight, or gram formula mass. Even though the atomic masses of the different elements in a periodic table are frequently decimals, the approximated full integer is utilized in computations for simplicity.
Formula to be used:
\[Molar{\text{ }}quantity = {\text{ }}\dfrac{{Given{\text{ }}mass}}{{{\text{ M}}olar{\text{ }}mass}}\]
Complete answer:
(i) The following rules are used to compute a compound's molecular mass:
1. The atomic masses of the elements that make up the specified compound must be known.
2. We multiply the atomic mass of a certain element by the number of instances of that element in the compound (written in the form of subscript of the element in the molecular formula).
3. We add all of the results together to get the molecular mass in grams/mole.
The molecular formula of lead nitrate is $Pb{(N{O_3})_2}$whose molar mass is calculated as $\{ 207.2 + 2 \times 14.01 + 6 \times 15.999\} g \cdot mo{l^{ - 1}}$
= $331.214gmo{l^{ - 1}}$
(ii) The molar quantity of a \[42.{\text{ }}65{\text{ }}g\]mass of lead nitrate is as follows:
\[Molar{\text{ }}quantity = {\text{ }}\dfrac{{Given{\text{ }}mass}}{{{\text{ M}}olar{\text{ }}mass}}\]
The given mass of the material is \[42.{\text{ }}65{\text{ }}g\]and the molar mass is $331.214gmo{l^{ - 1}}$. Putting these values, we get
\[Molar{\text{ }}quantity = {\text{ }}\dfrac{{42.65}}{{{\text{ 331}}{\text{.214}}}}\]
\[ = 0.128mol\]
Note:
It's worth mentioning that a compound's molar mass is known by a variety of common names. Among other names, the density of one mole of material in grams per mole is referred to as molecular mass, molecular weight, or gram formula mass. Even though the atomic masses of the different elements in a periodic table are frequently decimals, the approximated full integer is utilized in computations for simplicity.
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