What is the concentration created when 380.0g of Pb(NO3)2 is added to 330.0 mL of solution?
1 answer:
The concentration of lead nitrate is 3.48 M.
<u>Explanation:</u>
The molarity can be found by dividing moles of sucrose by its volume in litres. We can find the number of moles of sucrose by dividing the given mass by its molar mass. Now we can find the moles as,
Here mass of Pb(NO₃)₂ is 380 g
Molar mass of Pb(NO₃)₂ is 331.2 g/mol
Number of moles =
=
= 1.15 moles
Volume in Litres = 330 ml = 0.33 L
Molarity =
= 3.48 mol/L or 3.48 M
So the concentration of lead nitrate is 3.48 M.
You might be interested in
Answer:
Explanation:
Hello there!
In this case, we can divide the problem in three stages: (1) determine the empirical formula with the combustion analysis, (2) compute the molar mass of acid via the moles of the acid in the neutralization and (3) determine the molecular formula.
(1) In this case, since 8.58 g of carbon dioxide are released, we can first compute the moles of carbon in the compound:
And the moles of hydrogen due to the produced 1.50 grams of water:
Next, to compute the mass and moles of oxygen, we need to use the initial 3.4 g of the acid:
Thus, the subscripts in the empirical formula are:
As they cannot be fractions.
(2) In this case, since the acid is monoprotic, we can compute the moles by multiplying the concentration and volume of KOH:
Which are equal to the moles of the acid:
And the molar mass:
(3) Finally, since the molar mass of the empirical formula is:
7*12.01 + 6*1.01 + 2*16.00 = 122.13 g/mol
Thus, since the ratio of molar masses is 122.86/122.13 = 1, we infer that the empirical formula equals the molecular one:
Best regards!