Question

27. An organic compound is 38 compound? 15 pts C, 97% and 5161 W 26. How many molecules of ethane (CH) are present in 0.822 g of CHA T 29 Calculate the mass of NaCl in grams required to prepare 5.00 x 10 ml of 20 M

Answer 1

Assume there is 100 g of given compound ,then

Mass of C :: 38.66 g

Mass of H ;; 9.73 g

Mass of S:: 51.61 g

# mols of C = mass of C/molar mass of C = 38.66 g/12.0 g/mol
= 3.22 mols

# mols of H :: Mass of H/molar mass of H = 9.73 g/1.008 g/mol
= 9.65 mol

# mols os S :: Mas of S /molar mass of S = 51.61 g/32.065 g/mol
= 1.61 mols

DIvide each mols by smallest number to get simplest mole ratio

C ::3.22 mols/1.61 mols= =2
H ::9.65 mol/1.61 mols =6
S:: 1.61/1.61 = 1

therefore emperical formula of the compound = C2H6S
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we have molar mass of C2H6 = 2*12.01 g/mol + 6*1.01 g/mol
= 30.07 g/mol
given mass of ethane = 0.822 g
SO # moles of ethane present = given mass/molar mass
= 0.822 g/30.07g/mol = 0.02734 mols

we know 1 mole contains avogadro number of particles
0.02734 mols ethane contain = 0.02734 *6.022 x10^23 molecules =
= 0.1646 x10^23 moleculs of ethane
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GIven
molarity of solution = 2.50 M
volume of solution needed = 5 x10^2 ml = 0.5 Liters
So number of moles of solute [NaCL] required = molarity*volume of
solution in Liters= 2.5 M* 0.5 L
= 1.25 mols

Now we have number of moles of NaCL needed = 1.25 mols
Also we know molar mass of NaCl = 1*22.99 g/mol + 1*35.45 g/mol
= 58.44 g/mol
Therefore mass of NaCL needed = Number of moles of NaCl needed*molar mass
=1.25 mol* 58.44 g/mol = 73.05 g
mass of NaCl= 73.05 g
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