If $6 \cdot g$ $6g$ urea is dissolved in $1 \cdot L$ $1L$ of water, what is the resultant concentration?

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If $6 \cdot g$ $6g$ urea is dissolved in $1 \cdot L$ $1L$ of water, what is the resultant concentration?

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Answer 1 · 2017-07-01T10:00:53

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

Explanation:

And urea $O = C {(N H_{2})}_{2}$ $O=C(NH_2)_2$ has a molar mass of $60.06 \cdot g \cdot m o l^{- 1}$ $60.06*g*mol^-1$ .

And thus $[O = C {(N H_{2})}_{2}] = \frac{\frac{6 \cdot g}{60.06 \cdot g \cdot m o l^{- 1}}}{1 \cdot L} = 0.010 \cdot m o l \cdot L^{- 1}$ $[O=C(NH_2)_2]=((6*g)/(60.06*g*mol^-1))/(1*L)=0.010*mol*L^-1$ .

I presume the solvent is water......in which it has some considerable solubility.

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If $6 \cdot g$ $6g$ urea is dissolved in $1 \cdot L$ $1L$ of water, what is the resultant concentration?

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If $6 \cdot g$ $6g$ urea is dissolved in $1 \cdot L$ $1L$ of water, what is the resultant concentration?