<span>70.4 mg CO2 x 1.0 g /1000 mg x 1 mole CO2/ 44 gCO2 x 1 mole C/1 mole CO2 = 0.0016 moles C
14.4 mg H2O x 1.0 g/1000 mg x 1 mole H2O/18 g H2O x 2 moles H/ 1 mole H2O = 0.0016 moles O
molar mass of C=12 g/mole
molar mass of H=1 g/mole
0.0016 moles C x 12 g C/ 1 mole C = 0.0192 g C or 19.2 mg C
0.00156 moles H x 1 g H/1 mole H = 0.00156 g H or 1.56 mg H
mg O= 30.4 mg vanillin - 19.2 mg C – 1.56 mg H = 9.64 mg O
molar mass of O=16 g/mole
9.64 mg O x 1 g/1000 mg x 1 mole O/16.0 g = 0.000602
C.0016 H.0016 O.000602; divide all the moles by the smallest value of0.000602
C2.66H2.66O1 is the empirical formula;
to obtain whole numbers multiply by 3
3[C2.66H2.66O1] = C8H8O3
above formula weight: 8(C) + 8(H) + 3(O) = 8(12) + 8(1) + 3(16) = 152 amu
The empirical formula weight and the molecular formula weight are the same .
Molecular formula is C8H8O3.</span>
Answer is: 79.8 grams of copper(II) sulfate.
N(CuSO₄) = 3.01·10²³; number of molecules.
n(CuSO₄) = N(CuSO₄) ÷ Na.
n(CuSO₄) = 3.01·10²³ ÷ 6.02·10²³ 1/mol.
n(CuSO₄) = 0.5 mol; amount of substance.
m(CuSO₄) = n(CuSO₄) · M(CuSO₄).
m(CuSO₄) = 0.5 mol · 159.6 g/mol.
m(CuSO₄) = 79.8 g; mass of substance.
M - molar mass.
Answer:
34.9 g/mol is the molar mass for this solute
Explanation:
Formula for boiling point elevation: ΔT = Kb . m . i
ΔT = Temperatures 's difference between pure solvent and solution → 0.899°C
Kb = Ebullioscopic constant → 0.511°C/m
m = molality (moles of solute/1kg of solvent)
i = 2 → The solute is a strong electrolyte that ionizes into 2 ions
For example: AB ⇒ A⁺ + B⁻
Let's replace → 0.899°C = 0.511 °C/m . m . 2
0.899°C / 0.511 m/°C . 2 = m → 0.879 molal
This moles corresponds to 1 kg of solvent. Let's determine the molar mass
Molar mass (g/mol) → 30.76 g / 0.879 mol = 34.9 g/mol
V1 = 445ml V2 = 499ml
T1 = 274 K T2 = ?
By Charles Law,
V1/T1 = V2/T2
445/274 = 499/T2
By solving we get,
T2 = 307.25 K