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

21 \times 6.023 \times 10^{23}

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22.8 \times 6.023 \times 10^{23}

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200 \times 6.023 \times 10^{23}

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228 \times 6.023 \times 10^{23}

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Solution

The correct option is D $228 \times 6.023 \times 10^{23}$
$58.5 g of N a C l = \frac{58.5 g}{58.5 g/mol} = 1 mol of N a C l$

$360 g of H_{2} O = \frac{360 g}{18 g/mol} = 20 mol of H_{2} O$

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

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