How do we make a $1 \cdot m o l \cdot L^{- 1}$ $1molL^-1$ solution of ethanolamine, $H_{2} N C H_{2} C H_{2} O H$ $H_2NCH_2CH_2OH$ ?

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How do we make a $1 \cdot m o l \cdot L^{- 1}$ $1molL^-1$ solution of ethanolamine, $H_{2} N C H_{2} C H_{2} O H$ $H_2NCH_2CH_2OH$ ?

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Answer 1 · 2017-04-18T17:30:52

$Molarity$ $"Molarity"$ $=$ $=$ $\frac{Moles of solute}{Volume of solution}$ $"Moles of solute"/"Volume of solution"$

Explanation:

And so we take a $61.08 \cdot g$ $61.08*g$ mass of ethanolamine, and make it up to a $1 \cdot L$ $1*L$ volume in a volumetric flask:

$Molarity = \frac{\frac{61.08 \cdot g}{61.08 \cdot g \cdot m o l^{- 1}}}{1.00 \cdot L} = 1.00 \cdot m o l \cdot L^{- 1}$ $"Molarity"=((61.08*g)/(61.08*g*mol^-1))/(1.00*L)=1.00*mol*L^-1$

Would this give rise to a basic or acidic solution?

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How do we make a $1 \cdot m o l \cdot L^{- 1}$ $1molL^-1$ solution of ethanolamine, $H_{2} N C H_{2} C H_{2} O H$ $H_2NCH_2CH_2OH$ ?

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Explanation:

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How do we make a $1 \cdot m o l \cdot L^{- 1}$ $1molL^-1$ solution of ethanolamine, $H_{2} N C H_{2} C H_{2} O H$ $H_2NCH_2CH_2OH$ ?