Question

A solution contains $58.5\text{g}$ of common salt in $360\text{g}$ of water. Calculate the total number of protons in the solution.

• A. $21\times 6.023\times {10}^{23}$
• B. $22.8\times 6.023\times {10}^{23}$
• C. $200\times 6.023\times {10}^{23}$
• D. $228\times 6.023\times {10}^{23}$

Answer 1

The correct option is D $228\times 6.023\times {10}^{23}$

$58.5\text{g of}NaCl=\frac{58.5\text{g}}{58.5\text{g/mol}}=1\text{mol of}NaCl$

$360\text{g of}{H}_{2}O=\frac{360\text{g}}{18\text{g/mol}}=20\text{mol of}{H}_{2}O$

Number of electrons in 1 formula unit of $NaCl=11+17=28e^{-}$
Number of electrons in 1 molecule of $H_{2}O=2+8=10$
Number of electrons in 1 mol of $NaCl=28\times N_A$
Number of electrons in 20 mol of $H_{2}O=200\times N_A$
So, the total number of electrons in the solution = $228N_A$