What is the molality when 0.15 mol SO2 is dissolved in 100 ml water? Remember 1 ml of water = 1 gram of water)

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What is the molality when 0.15 mol SO2 is dissolved in 100 ml water? Remember 1 ml of water = 1 gram of water)

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Answer 1 · 2018-05-21T18:38:02

$Molality \equiv 1.5 \cdot m o l \cdot k g^{- 1}$ $"Molality"-=1.5*mol*kg^-1$ ...

Explanation:

By definition, $molality \equiv \frac{moles of solute}{kilograms of solvent}$ $"molality"-="moles of solute"/"kilograms of solvent"$

And so here... $m_{S O_{2}} = \frac{0.15 \cdot m o l}{100 \cdot m L \times 1 \cdot g \cdot m L^{- 1} \times 10^{- 3} k g \cdot g^{- 1}}$ $m_(SO_2)=(0.15*mol)/(100*mLxx1*g*mL^-1xx10^-3kg*g^-1)$

$\equiv 1.50 \cdot m o l \cdot k g^{- 1}$ $-=1.50*mol*kg^-1$ ...

...sometimes this solution is referred to as $sulfurous acid$ $"sulfurous acid"$ , i.e. $H_{2} S O_{3} \equiv H_{2} O \cdot S O_{2}$ $H_2SO_3-=H_2O*SO_2$ ...the smell is sulfurous but quite pleasant. Home brewers often use sulfurous acid to sanitize their carboys...

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What is the molality when 0.15 mol SO2 is dissolved in 100 ml water? Remember 1 ml of water = 1 gram of water)

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