Answer:
Molecular mass of X is 73g/mol
Explanation:
To answer this question you need to know that Freezing-point depression (A colligative property) is the decrease of the freezing point of a solvent on the addition of a non-volatile solute. The formula is:
ΔT = Kf m
Where ΔT is the freezing point depression (2,57°C - (-1,4°C) = 3,97°C Where 2,57°C is the melting point of formamide.
Kf that is freezing point molar constant of the solvent (4,25°C/m)
And m that is molality (moles of solute/kg of solvent). Replacing:
m = [tex]\frac{3,97}{4,25} =0,934 m[/tex]
Knowing you have 80,0g of solvent ≡ 0,0800 kg:
0,934 mol solute/kg solvent×0,08 kg solvent = 0,07472 moles of solute ≡ moles X
As grams of X are 5,42, molecular mass of X is:
[tex]\frac{5,42g}{0,07472 moles}[/tex] = 73 g/mol
I hope it helps!
Answer : The molar mass of X is 72 g/mol.
Explanation : Given,
Mass of X (solute) = 5.42 g
Mass of formamide (solvent) = 80.0 g = 0.080 kg
Molar mass of formamide = 45.04 g/mole
Formula used :
[tex]\Delta T_f=i\times K_f\times m\\\\T^o-T_s=i\times K_f\times\frac{\text{Mass of compound X}}{\text{Molar mass of compound X}\times \text{Mass of formamide in Kg}}[/tex]
where,
[tex]\Delta T_f[/tex] = change in freezing point
[tex]\Delta T_s[/tex] = freezing point of solution = [tex]-1.4^oC[/tex]
[tex]\Delta T^o[/tex] = freezing point of formamide = [tex]2.2^oC[/tex]
i = Van't Hoff factor = 1 (for no-electrolyte)
[tex]K_f[/tex] = freezing point constant for formamide = [tex]3.85^oC/m[/tex]
m = molality
Now put all the given values in this formula, we get
[tex][2.2-(-1.4)]^oC=1\times (3.85^oC/m)\times \frac{5.42g}{\text{Molar mass of compound X}\times 0.080kg}[/tex]
[tex]\text{Molar mass of compound X}=72.4g/mol\approx 72g/mol[/tex]
Therefore, the molar mass of X is 72 g/mol.
