Find out the number of carbon atoms present i | vedclass

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

Question

(C) The chemical formula of glucose is $C_6H_{12}O_6$. One mole of glucose contains $6 \ mol$ of carbon atoms. Therefore,$0.35 \ mol$ of glucose conta

Vedclass · Accepted Answer

$1.264 	imes 10^{24}$

Vedclass · Answer

(C) The chemical formula of glucose is $C_6H_{12}O_6$. One mole of glucose contains $6 \ mol$ of carbon atoms. Therefore,$0.35 \ mol$ of glucose contains $0.35 	imes 6 = 2.1 \ mol$ of carbon atoms. The number of carbon atoms is calculated as $2.1 	imes N_A$,where $N_A = 6.022 	imes 10^{23} \ mol^{-1}$. Number of carbon atoms $= 2.1 	imes 6.022 	imes 10^{23} = 12.6462 	imes 10^{23} = 1.26462 	imes 10^{24}$. Thus,the correct option is $C$.

Find out the number of carbon atoms present in $0.35 \ mol$ of glucose.

Similar Questions

What is the number of hydrogen atoms present in $5.4 \ g$ of urea?

The number of $Cl^{-}$ ions in $333 \ g$ of anhydrous $CaCl_2$ will be ................ $N_A$.

How many protons are present in $1.8 \ g$ of $NH_{4}^{+}$?

Find the number of electrons present in $34 \ g$ of $NH_{3(g)}$. (Express your answer in terms of $N_A$)

The number of atoms in $8 \ g$ of sodium is $x \times 10^{23}$. The value of $x$ is ...... (Nearest integer).
$[ \text{Given}: N_{A} = 6.02 \times 10^{23} \ mol^{-1}, \text{Atomic mass of } Na = 23.0 \ u ]$

Vedclass Test Series

Exam Paper Generator

Online Exam Module