Answer
405k+ views
Hint: In the question, molarity of the lead nitrate solution i.e. 1 is given. Recall the formula to calculate the molarity of the solution. In the formula of molarity, volume must be in litres, but we give volume in millilitres. You also need to find out the moles of lead nitrate solution. Given that molecular mass of lead nitrate is 331 g.
Complete Solution :
- We are given 1.00 molar solution of lead nitrate i.e. molarity of the solution is given.
Molarity of a solution, denoted by M, is defined as the number of moles of the solute present in 1 litre of the solution. Thus, formula for calculating molarity is:
Molarity (M) = $\dfrac{{{\text{Number of moles of solute}}}}{{{\text{Volume of solution in litres}}}}$
- We are also given, volume of the solution = 100 millilitres.
100 mL = 0.1 litres.
- We are also required to calculate the number of moles of solute that is, $Pb{(N{O_3})_2}$ here to use in the formula of molarity.
Formula for calculating number of moles is:
Number of moles = $\dfrac{{{\text{Given mass}}}}{{{\text{Molar mass}}}}$
- Now, let the mass of $Pb{(N{O_3})_2}$ required to make 100 millimetres of a 1.00 molar solution is $x$.
And, molar mass of $Pb{(N{O_3})_2}$ = 331 grams (given).
Thus, number of moles of lead nitrate, $Pb{(N{O_3})_2}$= $\dfrac{x}{{331}}$
Now, molarity of $Pb{(N{O_3})_2}$ = $\dfrac{{{\text{Moles of }}Pb{{(N{O_3})}_2}}}{{{\text{Volume of the solution (in L)}}}}$
$ \Rightarrow 1.00 = \dfrac{x}{{331 \times 0.1}}$
$ \Rightarrow x = 33.1$ grams.
Hence, mass of lead nitrate i.e. $Pb{(N{O_3})_2}$ needed to make 100 millimetres of a 1.00 molar solution is 33.1 grams.
So, the correct answer is “Option B”.
Note: You can also be given molality which is denoted by m. It is defined as the number of moles of solute present in 1 kg of solvent. Thus,
Molality (m) = $\dfrac{{{\text{No}}{\text{. of moles of solute}}}}{{{\text{Mass of solvent (in kg)}}}}$
Complete Solution :
- We are given 1.00 molar solution of lead nitrate i.e. molarity of the solution is given.
Molarity of a solution, denoted by M, is defined as the number of moles of the solute present in 1 litre of the solution. Thus, formula for calculating molarity is:
Molarity (M) = $\dfrac{{{\text{Number of moles of solute}}}}{{{\text{Volume of solution in litres}}}}$
- We are also given, volume of the solution = 100 millilitres.
100 mL = 0.1 litres.
- We are also required to calculate the number of moles of solute that is, $Pb{(N{O_3})_2}$ here to use in the formula of molarity.
Formula for calculating number of moles is:
Number of moles = $\dfrac{{{\text{Given mass}}}}{{{\text{Molar mass}}}}$
- Now, let the mass of $Pb{(N{O_3})_2}$ required to make 100 millimetres of a 1.00 molar solution is $x$.
And, molar mass of $Pb{(N{O_3})_2}$ = 331 grams (given).
Thus, number of moles of lead nitrate, $Pb{(N{O_3})_2}$= $\dfrac{x}{{331}}$
Now, molarity of $Pb{(N{O_3})_2}$ = $\dfrac{{{\text{Moles of }}Pb{{(N{O_3})}_2}}}{{{\text{Volume of the solution (in L)}}}}$
$ \Rightarrow 1.00 = \dfrac{x}{{331 \times 0.1}}$
$ \Rightarrow x = 33.1$ grams.
Hence, mass of lead nitrate i.e. $Pb{(N{O_3})_2}$ needed to make 100 millimetres of a 1.00 molar solution is 33.1 grams.
So, the correct answer is “Option B”.
Note: You can also be given molality which is denoted by m. It is defined as the number of moles of solute present in 1 kg of solvent. Thus,
Molality (m) = $\dfrac{{{\text{No}}{\text{. of moles of solute}}}}{{{\text{Mass of solvent (in kg)}}}}$
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Why Are Noble Gases NonReactive class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
At which age domestication of animals started A Neolithic class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write a letter to the principal requesting him to grant class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)