Question

Compute the value of $X;x=9.4g$ of phenol $\left(C_6{H}_{5}OH\right)+6.02\times {10}^{23}$ molecule of phenol $-0.2$ moles of phenol. Here $x$ is:

• A. $0.9mole$
• B. $9.2g$
• C. $0.1mole$
• D. $6.02\times {10}^{23}mole$

Answer 1

The correct option is A $0.9mole$

Moalr mass of $C_6{H}_{5}OH=6\times 12+6\times 1+16=94$ g

No of moles of 9.4 g phenol $=\frac{9.4}{94}=0.1$

No. of moles of $6.022\times {10}^{23}$ molecule $=\frac{6.022\times {10}^{23}}{6.022\times {10}^{23}}=1$

x$=0.1+1-0.2=0.9$ mole