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Answer:
99.24%.
Explanation:
- NaCl reacted with AgNO₃ as in the balanced equation:
<em>NaCl + AgNO₃ → AgCl(↓) + NaNO₃,</em>
1.0 mol of NaCl reacts with 1.0 mol of AgNO₃ to produce 1.0 mol of AgCl and 1.0 mol of NaNO₃.
- We need to calculate the no. of moles of AgCl produced:
no. of moles of AgCl = mass/molar mass = (2.044 g)/(143.32 g/mol) = 0.0143 mol.
- Now, we can calculate the no. of moles of NaCl that can precipitated as AgCl (0.0143 mol), these moles represents the no. of moles of pure NaCl in the sample:
<em>using cross multiplication:</em>
1.0 mol of NaCl produce → 1.0 mol of AgCl, from the stichiometry.
∴ 0.0143 mol of NaCl produce → 0.0143 mol of AgCl.
- Now, we can get the mass of puree NaCl in the sample:
mass of pure NaCl = (no. of moles of pure NaCl)(molar mass of NaCl) = (0.0143 mol)(58.44 g/mol) = 0.8357 g.
∴ The percentage of NaCl in the impure sample = [(mass of pure NaCl)/(mass of the impure sample)] x 100 = [(0.8357 g)/(0.8421 g)] x 100 = 99.24%.
Answer:
616,0 ng is the right answer.
Explanation:
You should know that 1 mole = 1 .10^9 nanomoles
Get the rule of three.
1 .10^9 nanomoles ...................... 56.0 gr
11 nanomoles .....................
(11 x 56) / 1 .10^9 nanomoles = 6.16 x 10^-7 gr
Let's convert
6.16 x 10^-7 gr x 1 .10^9 = 616 ngr
