Question 1.11
NCERT Class XI Chemistry
What is the concentration of sugar (C12H22O11) in mol L-1 if its 20 g are dissolved in enough water to make a final volume up to 2L?
Hint and Answer
(detailed solution after video)
Basically we have to calculate the mols of sugar in 20 grams and divide that by the given volume of 2 liters to obtain the final answer of 0.03 mol/liter.
The significant figures in the final answer will be obtained according to the rules of addition and division.
We will start by calculating the molar mass of sugar first. The details have been given in the detailed solution below.
Video Explanation
(detailed solution given after this video)
Please watch this youtube video for a quick explanation of the solution:
Solution in Detail
$ \begin{aligned} &\underline{\text{Molar Mass of Sugar}}\\ &\text{atomic masses of C, H and O:}\\ &\text{ C = 12.00u, H = 1.008u, O= 16.00u}\\ &\therefore \text{molar mass of C}_{12}\text{H}_{22}\text{O}_{11}\text{is:} \\ &=(12.00 \times 12) + (1.008 \times 22) + (16.00 \times 11)\\ &= 144.00 + 22.176 + 176.00 \\ &= 342.18\text{g/mol}\\\\ &\underline{\text{Moles of Sugar in } 20 \text{ grams}}\\\\ &=\frac{\text{given mass}}{\text{molar mass}} = \frac{20}{342.18} \text{. . . (1)}\\\\ &\underline{\text{conc. of Sugar in } 2 \text{ liters}}\\\\ &=\frac{\text{moles of sugar}}{\text{volume of soln.}}\\\\ &=\frac{(20/342.18) \text{ mols}}{2 \text{ liters}} = 0.0292\\\\ &=0.03 \text{ mol/l (rounded to 1 SF)}\:\underline{Ans}\\ &\text{(SF in 2 liters and 20 grams were 1)}\\ \end{aligned} $
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