A person adds 1.71 gram of sugar ((C\_{12}H\_{22}O\_{11})) i... | Chemistry | SolveeIt | Solveeit

Today, August 18

Question

A person adds 1.71 gram of sugar $(C_{12}H_{22}O_{11})$ in order to sweeten his tea. The number of carbon atoms added are (mol. mass of sugar = 342)

A

$3.6 \times 10^{22}$

B

$7.2 \times 10^{21}$

C

0.05

D

$6.6 \times 10^{22}$

Answer

$3.6 \times 10^{22}$

Explanation

Solution

Moles of sugar added $=\frac{1.71}{342}=5\times10^{-3}$ Carbon atoms added $=12\times5\times10^{-3}\times6.02\times10^{23}$ $=3.61\times10^{22}$