Na = 23 x 2.40 = 55.2
O = 16 x 2.40 = 38.4
H = 1 x 2.40 = 2.40
55.2 + 38.4 + 2.4 = 96
2.40 mol of NaOH = 96 amu
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!
Complete question is;
A drop of water has a volume of approximately 7 × 10⁻² ml. How many water molecules does it contain? The density of water is 1.0 g/cm³.
This question will require us to first find the number of moles and then use avogadro's number to get the number of water molecules.
<em><u>Number of water molecules = 2.34 × 10²¹ molecules</u></em>
We are given;
Volume of water; V = 7 × 10⁻² ml
Density of water; ρ = 1 g/cm³ = 1 g/ml
Formula for mass is; m = ρV
m = 1 × 7 × 10⁻²
m = 7 × 10⁻² g
from online calculation, molar mass of water = 18.01 g/mol
Number of moles(n) = mass/molar mass
Thus;
n = (7 × 10⁻²)/18.01
n = 3.887 × 10⁻³ mol
from avogadro's number, we know that;
1 mol = 6.022 × 10²³ molecules
Thus,3.887 × 10⁻³ mol will give; 6.022 × 10²³ × 3.887 × 10⁻³ = 2.34 × 10²¹ molecules
Read more at; brainly.in/question/17990661
