The mass of hydrogen (in grams) present in 1. | vedclass

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(A) Density of water = $1.0 	ext{ g cm}^{-3}$.Volume of water = $1.0 	ext{ L} = 1000 	ext{ mL}$.Mass of $1.0 	ext{ L}$ of water = $1000 	ext{ mL}

Vedclass · Accepted Answer

$1.11 	imes 10^2$

Vedclass · Answer

(A) Density of water = $1.0 	ext{ g cm}^{-3}$.Volume of water = $1.0 	ext{ L} = 1000 	ext{ mL}$.Mass of $1.0 	ext{ L}$ of water = $1000 	ext{ mL} 	imes 1.0 	ext{ g mL}^{-1} = 1000 	ext{ g}$.Molar mass of $H_2O = (2 	imes 1.008) + 16.00 = 18.016 	ext{ g mol}^{-1} \approx 18 	ext{ g mol}^{-1}$.In $18 	ext{ g}$ of $H_2O$,the mass of hydrogen is $2 	ext{ g}$.Therefore,the mass of hydrogen in $1000 	ext{ g}$ of $H_2O = (2 	ext{ g} / 18 	ext{ g}) 	imes 1000 	ext{ g} = 111.11 	ext{ g}$.This is equal to $1.11 	imes 10^2 	ext{ g}$.

The mass of hydrogen (in grams) present in $1.0 \text{ L}$ of pure water of density $1.0 \text{ g cm}^{-3}$ is

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