A 20.0-g sample of sugar is dissolved in 59.8 g water. What is the concentration of the solution?

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A 20.0-g sample of sugar is dissolved in 59.8 g water. What is the concentration of the solution?

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Answer 1 · 2017-06-10T20:31:54

Approx. $1.9 \cdot m o l \cdot L^{- 1}$ $1.9*mol*L^-1$ with respect to glucose........ $C_{6} H_{12} O_{6}$ $C_6H_12O_6$ .....

Explanation:

With aqueous solutions it is usual to assume that dissolution of the solute occurs with little to NO change in solution volume.

And thus $Concentration = \frac{Moles of solute}{Volume of solution.}$ $"Concentration"="Moles of solute"/"Volume of solution."$

$= \frac{\frac{20.0 \cdot g}{180.16 \cdot g \cdot m o l^{- 1}}}{\frac{59.8 \cdot g}{1.00 \cdot g \cdot m L^{- 1}} \times 10^{- 3} \cdot L \cdot m L^{- 1}} = ? ? \cdot m o l \cdot L^{- 1}$ $=((20.0*g)/(180.16*g*mol^-1))/((59.8*g)/(1.00*g*mL^-1)xx10^-3*L*mL^-1)=??*mol*L^-1$ .

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A 20.0-g sample of sugar is dissolved in 59.8 g water. What is the concentration of the solution?

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