For the answer to the question above, use these formulas in solving your problem and as a guide.
<span>
MM = 342 (g/mol) </span>
<span>171 (g) / 342(g/mol) = x mol of sucrose </span>
<span>x moles of sucrose/ 1.25 L = Molarity of soultion
</span>I hope I helped you with your problem. Have a beautiful day!
For the first part, use the question M=mol/vol (liters)
To do this, you have the given 1.6 M solution
divide the 360g by the molar mass of ethanol (44.07) to get moles
360/44.07=8.16 mol
so
1.6M = 8.16 mol/x vol
volume: 5.1 Liters
Answer:
34,6g of (NH₄)₂SO₄
Explanation:
The boiling-point elevation describes the phenomenon in which the boiling point of a liquid increases with the addition of a compound. The formula is:
ΔT = kb×m
Where ΔT is Tsolution - T solvent; kb is ebullioscopic constant and m is molality of ions in solution.
For the problem:
ΔT = 109,7°C-108,3°C = 1,4°C
kb = 1.07 °C kg/mol
Solving:
m = 1,31 mol/kg
As mass of X = 600g = 0,600kg:
1,31mol/kg×0,600kg = 0,785 moles of ions. As (NH₄)₂SO₄ has three ions:
0,785 moles of ions×
= 0,262 moles of (NH₄)₂SO₄
As molar mass of (NH₄)₂SO₄ is 132,14g/mol:
0,262 moles of (NH₄)₂SO₄×
= <em>34,6g of (NH₄)₂SO₄</em>
<em></em>
I hope it helps!
Answer:
Normalidad = 4N
%p/V = 27.6%
Explanation:
La solución 2M de carbonato de potasio contiene 2moles de carbonato por litro de solución. La normalidad son los equivalente de carbonato de potasio (2eq/mol) por litro de solución:
2moles * (2eq/mol) = 4eq / 1L = 4N
El porcentaje peso volumen es el peso de carbonato en gramos dividido en el volumen en mL por 100:
%p/V:
Masa K2CO3 -Masa molar: 138.205g/mol-
2moles * (138.205g/mol) = 276g K2CO3
Volumen:
1L * (1000mL/1L) = 1000mL
%p/V:
276g K2CO3 / 1000mL * 100
<h3>%p/V = 27.6%</h3>
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
