What would you need to do to calculate the molality of 10 mol of NaCl in 200
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The relative molecular mass of acid A : 50 g/mol
<h3>Further explanation</h3>Given
40.0 cm³(40 ml) of 0.2M sodium hydroxide
0.2g of a dibasic acid
Required
the relative molecular mass of acid A
Solution
Titration formula
M₁V₁n₁=M₂V₂n₂
n=acid/base valence(number of H⁺/OH⁻)
NaOH ⇒ n = 1
Dibasic acid = diprotic acid (such as H₂SO₄)⇒ n = 2
mol = M x V
Input the value in the formula :(1 = NaOH, 2=dibasic acid)
0.2 x 40 x 1 = M₂ x V₂ x 2
M₂ x V₂ = 4 mlmol = 4.10⁻³ mol ⇒ mol of Acid A
The relative molecular mass of acid A (M) :
We can use combined gas laws to solve for the volume of the gas
where P - pressure, V - volume , T - temperature and k - constant
parameters for the first instance are on the left side and parameters for the second instance are on the right side of the equation
T1 - temperature in Kelvin - 20 °C + 273 = 293 K
T2 - 40 °C + 273 = 313 K
substituting the values
V = 17.8 L
volume of the gas is 17.8 L
where P - pressure, V - volume , T - temperature and k - constant
parameters for the first instance are on the left side and parameters for the second instance are on the right side of the equation
T1 - temperature in Kelvin - 20 °C + 273 = 293 K
T2 - 40 °C + 273 = 313 K
substituting the values
V = 17.8 L
volume of the gas is 17.8 L