Among 10^{-9} g (each) of the following elem | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The number of atoms is calculated using the formula: $	ext{Number of atoms} = \frac{	ext{Mass}}{	ext{Molar Mass}} 	imes N_A$.Since the mass is

Vedclass · Accepted Answer

$Pr$

Vedclass · Answer

(B) The number of atoms is calculated using the formula: $	ext{Number of atoms} = \frac{	ext{Mass}}{	ext{Molar Mass}} 	imes N_A$.Since the mass is constant $(10^{-9} \ g)$ for all elements,the number of atoms is inversely proportional to the molar mass.Therefore,the element with the lowest molar mass will have the highest number of atoms.The molar masses are:$\bullet M_{Po} = 209 \ g/mol$$\bullet M_{Pr} = 141 \ g/mol$$\bullet M_{Pb} = 207 \ g/mol$$\bullet M_{Pt} = 195 \ g/mol$Comparing these values,$Pr$ has the lowest molar mass $(141 \ g/mol)$.Thus,$Pr$ will have the highest number of atoms.

Among $10^{-9} \ g$ (each) of the following elements,which one will have the highest number of atoms?
Element: $Pb, Po, Pr$ and $Pt$

Similar Questions

The oxide of a metal has the molecular formula $M_2O_3$. If the equivalent mass of the metal is $9$,what is the molecular mass of the oxide?

An appropriate method for molecular weight determination of chloroform $(CHCl_3)$ is:

What is the average atomic mass of chlorine if its two isotopes $^{35}Cl$ and $^{37}Cl$ exist in a relative abundance of $75\%$ and $25\%$ respectively?

The equivalent weight of a divalent metal is $31.82$. The weight of one atom of this metal is:

Equivalent weights of $X_2Y$ and $X_2Y_3$ are $38$ and $18$ respectively. Find the atomic masses of $X$ and $Y$:-

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Among $10^{-9} \ g$ (each) of the following elements,which one will have the highest number of atoms?Element: $Pb, Po, Pr$ and $Pt$