# please answer my question as soon as possible. i have my exams tomorrowa)Combustion of 0.2000g of vitamin C gives 0.2998g of CO2and 0.819g of H2O. What is the empirical formula of vitamin C?b)4 litres of water are added to 2L of 6 molar HCl solutions.What is the molarity of resulting solution?

a)
Mass of water produced in this reaction must be 0.0819g and not 0.819g.
Given that vitamin C on combustion gives carbon dioxide and water.Let vitamin C is X.
X (s) + O2(g) $\to$CO2​ (g) + H2O (l)
0.2000 g           0.2998g     0.0819g

Molar mass of CO2 = 12 + 2$×$16= 44 g
Amount of carbon in 44 g CO2 = 12 g
Amount of carbon in 0.2998g  CO2 = $\frac{12}{44}×0.2998$ g​ =0.0818 g

Molar mass of H2O = 1$×$2 + 16= 18 g
Amount of hydrogen in 18 g
H2O = 2 g
Amount of hydrogen  in 0.0819 g H2O​ = $\frac{2}{18}×0.0819$ g​ =0.0091 g

Percentage of carbon in fuel gas = $\frac{0.0818}{0.2000}×100$ = 40.9 %

Percentage of hydrogen in compound = $\frac{0.0091}{0.2000}×100\phantom{\rule{0ex}{0ex}}$ = 4.55 %​

Thus Percentage of oxygen in the compound= 100 - (40.9 + 4.55) = 54.55 %

 Element Atomic mass of element Percentage Moles of the constituents Molar ratio Simple whole number ratio C 12 40.9 $\frac{40.9}{12}=3.408$ $\frac{3.408}{3.408}=1$ $1×3=3$ H 1 4.55 $\frac{4.55}{1}=4.55$ $\frac{4.55}{3.408}=1.33$ $1.33×3=4$ O 16 54.55 $\frac{54.55}{16}=3.41$ $\frac{3.41}{3.408}=1.00$ $1×3=3$

Thus the simple ratio of C:H:O is 3:4:3

Thus empirical formula of the compound = C3H4O3

b)

Thus now the molarity of the solution is 2 M

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