Question

3.42 g of sucrose are dissolved in 18 g of water in a beaker. The number of oxygen atoms in the solution are

• A. 6-68 x ${10}^{23}$
• B. 6-09 x ${10}^{22}$
• C. 6-022 x ${10}^{23}$
• D. 6-022 x ${10}^{21}$

Answer 1

The correct option is A

6-68 x ${10}^{23}$

As we Know

• Number of moles of sucrose in 3.42 g=$\frac{3.42g}{3.42gmo{l}^{-1}}=0.01mol$
• Number of oxygen atoms in 1 mole of sucrose=${\mathrm{C}}_{12}{\mathrm{H}}_{22}{\mathrm{O}}_{11}$$\times N_A$
• Number of oxygen atoms in 0.01 mole of sucrose=0.01$\times 11\times N_A$
• Number of moles of ${\mathrm{H}}_2\mathrm{O}$ in 18 g =$\frac{18g}{18gmo{l}^{-1}}$=1
• No. of oxygen atoms in 1 mole of water =${\mathrm{N}}_{\mathrm{A}}$
• Total no. of oxygen atoms = (0.11 + 1) = 1-11 NA

=$1.11\times 6.022\times {10}^{23}$

=$6.68$$\times {10}^{23}$