Answer:
1.62
Explanation:
From the given information:
number of moles of benzamide 
= 0.58 mole
The molality = 
= 0.6837
Using the formula:
where;
dT = freezing point = 27
l = Van't Hoff factor = 1
kf = freezing constant of the solvent
∴
2.7 °C = 1 × kf × 0.6837 m
kf = 2.7 °C/ 0.6837m
kf = 3.949 °C/m
number of moles of NH4Cl = 
= 1.316 mol
The molality = 
= 1.5484
Thus;
the above kf value is used in determining the Van't Hoff factor for NH4Cl
i.e.
9.9 = l × 3.949 × 1.5484 m
l = 1.62
<span>2.40 - 1.68 =0.72 g of oxigen
moles = 0.72/16 g/mol=0.045
moles x = 1.68/ 55.9=0.03
0.03/0.03 = 1 = x
0.045 / 0.03 = 1.5 = O
to get whole numbers multiply by 2
x2O3
X2O3 +3 CO = 2 X + 3 CO2</span>
The volume (in mL) of calcium hydroxide, Ca(OH)₂ needed for the reaction is 19.8 mL
<h3>Balanced equation </h3>
2HCl + Ca(OH)₂ —> CaCl₂ + 2H₂O
From the balanced equation above,
- The mole ratio of the acid, HCl (nA) = 2
- The mole ratio of the base, Ca(OH)₂ (nB) = 1
<h3>How to determine the volume of Ca(OH)₂ </h3>
- Molarity of base, Ca(OH)₂ (Mb) = 1.48 M
- Volume of acid, HCl (Va) = 36 mL
- Molarity of acid, HCl (Ma) = 1.63 M
- Volume of base, Ca(OH)₂ (Vb) =?
MaVa / MbVb = nA / nB
(1.63 × 36) / (1.48 × Vb) = 2
58.68 / (1.48 × Vb) = 2
Cross multiply
2 × 1.48 × Vb = 58.68
2.96 × Vb = 58.68
Divide both side by 2.96
Vb = 58.68 / 2.96
Vb = 19.8 mL
Learn more about titration:
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