Answer:
The mass of 10 cm³of a 0.4 g/dm³ solution of sodium carbonate is 0.004 grams
Explanation:
The question is with regards to density calculations
The density of the given sodium carbonate solution, ρ = 0.4 g/dm³
The volume of the given solution of sodium carbonate, V = 10 cm³ = 0.01 dm³


Therefore, we have;

The mass, "m", of the sodium carbonate in = ρ×V = 0.4 g/dm³ × 0.01 dm³ = 0.004 g
The mass of 10 cm³ (10 cm³ = 0.01 dm³) of a 0.4 g/dm³ solution of sodium carbonate, m = 0.004 g.
V ( HCl ) = 45.00 mL in liters : 45.00 / 1000 => 0.045 L
M ( HCl ) = ?
V ( NaOH ) = 25.00 / 1000 => 0.025 L
M ( NaOH) = 0.2000 M
number of moles NaOH :
n = M x V = 0.2000 x 0.025 => 0.005 moles of NaOH
Mole ratio:
HCl + NaOH = NaCl + H2O
1 mole HCl ---------- 1 mole NaOH
? mole HCl ---------- 0.005 moles NaOH
moles HCl = 0.005 x 1 / 1
= 0.005 moles of HCl :
M ( HCl ) = n / V
M ( HCl ) = 0.005 / 0.045
= 0.1111 M
hope this helps!
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