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2 years ago

What volume of 0.4567 m h2so4 is required to neutralize 30.00 ml of 0.3210 m naoh?

1 answer:

3 0

Answer is: volume of H₂SO₄ is 42.1 mL.
Chemical reaction: H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O.
c(H₂SO₄) = 0,4567 M = 0,4567 mol/L.
V(NaOH) = 30 mL ÷ 1000 mL/L = 0,03 L.
c(NaOH) = 0,321 M = 0,321 mol/L.
n(NaOH) = c(NaOH) · V(NaOH).
n(NaOH) = 0,321 mol/L · 0,030 L.
n(NaOH) = 0,00963 mol.
From chemical reaction: n(H₂SO₄) : n(NaOH) = 1 : 2.
n(H₂SO₄) = 0,01926 mol.
V(H₂SO₄) = n(H₂SO₄) ÷ c(H₂SO₄).
V(H₂SO₄) = 0,01926 mol ÷ 0,4567 mol/L.
V(H₂SO₄) = 0,0421 L = 42,1 mL.

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So, mole fraction of $C_2H_4$ = 0.3

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$\text{Average molecular weight of mixture}=\frac{_{i=1}^n\sum{\chi_im_i}}{n_i}$

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$n_i$ = number of observations

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$\text{Average molecular weight}=\frac{(\chi_{CH_4}\times M_{CH_4})+(\chi_{C_2H_4}\times M_{C_2H_4})+(\chi_{C_2H_2}\times M_{C_2H_2})+(\chi_{C_2H_2O}\times M_{C_2H_2O})}{4}$

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