What is the concentration of sugar $(C_{12} H_{22} O_{11})$ in mol $L^{- 1}$ if its $20 g$ are dissolved in enough water to make a final volume up to $2 L ?$

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Solution

Concentration of sugar

Given, 𝑀𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 $= 20 g$
$Molarmassofsolute (C_{12} H_{22} O_{11}) = 12 \times 12 + 22 \times 1 + 11 \times 16 = 342 g / m o l$

𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 $(i n L) = 2$
We know,

$Molarity(M) = \frac{Number of moles of solute}{Volume of solution in liters}$

$= \frac{\frac{Massofsugar}{Molarmassofsugar}}{2 L}$

$= \frac{\frac{20 g}{342 g}}{2 L}$

$= 0.02925 m o l L^{- 1}$

Final answer: 𝑀𝑜𝑙𝑎𝑟 𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛 $= 0.02925 m o l L^{- 1}$

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