Answer is: mass of methanol is 528.32 grams.
N(CH₃OH) = 9.93·10²⁴; number of methanol molecules.
n(CH₃OH) = N(CH₃OH) ÷ Na (Avogadro constant).
n(CH₃OH) = 9.93·10²⁴ ÷ 6.022·10²³ 1/mol.
n(CH₃OH) = 16.49 mol; amount of substance.
m(CH₃OH) = n(CH₃OH) · M(CH₃OH).
m(CH₃OH) = 16.49 mol · 32.04 g/mol.
m(CH₃OH) = 528.32 g.
Answer:
0.238 M
Explanation:
A 17.00 mL sample of the dilute solution was found to contain 0.220 M ClO₃⁻(aq). The concentration is an intensive property, so the concentration in the 52.00 mL is also 0.220 M ClO₃⁻(aq). We can find the initial concentration of ClO₃⁻ using the dilution rule.
C₁.V₁ = C₂.V₂
C₁ × 24.00 mL = 0.220 M × 52.00 mL
C₁ = 0.477 M
The concentration of Pb(ClO₃)₂ is:

A: making s sandcastle. This is because water and sand is only a mixture, so they do not react with each other. All the rest include chemical reactions!
Answer : The mole fraction of NaCl in a mixture is, 0.360
Explanation : Given,
Moles of
= 7.21 mole
Moles of
= 9.37 mole
Moles of
= 3.42 mole
Now we have to calculate the mole fraction of
.

Now put all the given values in this formula, we get:

Therefore, the mole fraction of NaCl in a mixture is, 0.360