An element,X has the following isotopic compo | vedclass

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

Question

(D) The weighted average atomic mass is calculated as:$	ext{Average atomic mass} = (\% 	ext{ abundance of isotope } 1 	imes 	ext{mass of isotope }

Vedclass · Accepted Answer

$200$

Vedclass · Answer

(D) The weighted average atomic mass is calculated as:$	ext{Average atomic mass} = (\% 	ext{ abundance of isotope } 1 	imes 	ext{mass of isotope } 1) + (\% 	ext{ abundance of isotope } 2 	imes 	ext{mass of isotope } 2) + (\% 	ext{ abundance of isotope } 3 	imes 	ext{mass of isotope } 3)$$	ext{Average atomic mass} = (0.90 	imes 200) + (0.08 	imes 199) + (0.02 	imes 202)$$	ext{Average atomic mass} = 180.00 + 15.92 + 4.04 = 199.96 \ amu$Rounding to the nearest whole number,the value is $200 \ amu$.

An element,$X$ has the following isotopic composition:
$^{200}X: 90\%$,$^{199}X: 8.0\%$,$^{202}X: 2.0\%$
The weighted average atomic mass of the naturally occurring element $X$ is closest to......$amu$.

Similar Questions

Naturally occurring boron has $20\%$ $_5B^{10}$ and $80\%$ $_5B^{11}$. The average atomic weight of boron is

What is the mass in gram of $1$ atom of an element if its atomic mass is $10 \ u$?

$10 \text{ g}$ of a metal $(M)$ reacts with oxygen to form $11.6 \text{ g}$ of oxide. What is the equivalent weight of $M$?

Calculate the molecular mass of a glucose $C_{6}H_{12}O_{6}$ molecule.

At $STP$,if $5.6 \ L$ of a gas weighs $7.5 \ g$,then identify the gas.

Vedclass Test Series

Exam Paper Generator

Online Exam Module

An element,$X$ has the following isotopic composition:$^{200}X: 90\%$,$^{199}X: 8.0\%$,$^{202}X: 2.0\%$The weighted average atomic mass of the naturally occurring element $X$ is closest to......$amu$.