Answer:
the quantity required can go from 117 ml (for maximum concentration) up to 2900 ml ( if the concentrated solution has molarity =0.420 M)
Explanation:
the amount of water required to dilute a solution V₁ liters of Molarity M₁ to V₂ liters of M₂
moles of hydrochloric acid = M₁ * V₁= M₂ * V₂
V₁ = V₂ * M₂/M₁
where
M₂ = 0.420 M
V₂ =2.90 L
Since the hydrochloric acid can be concentrated up to 38% p/V ( higher concentrations are possible but the evaporation rate is so high that handling and storage require extra precautions, like cooling and pressurisation)
maximum M₁ =38% p/V = 38 gr/ 0.1 L / 36.5 gr/mol = 10.41 M
then
min V₁ = V₂ * M₂/ max M₁ = 2.90 L* 0.420 M/ 10.41 M= 0.117 L = 117 ml
then the quantity required can go from 117 ml up to 2900 ml ( if M₁ = M₂)
Answer:
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Explanation:
Answer: 1.2 M
Explanation:
Given that:
volume of NaCl = 392 mL
Convert volume in milliliters to liters
(since 1000 mL = 1L
392 mL = 392/10000 = 0.392 L)
Concentration of NaCl = ?
Amount of moles of NaCl = 0.47 moles
Recall that the concentration of a solution depends on the amount of solute dissolved in a particular volume of solvent.
I.e Concentration in mol/L =
Amount in moles / Volume in liters
= 0.47 moles / 0.392 L
= 1.199 mol/L (Round up to the nearest tenth as 1.2 M)
Note that molarity is the same as concentration in moles per litres.
Thus, the molarity of 392 mL of solution that contains 0.47 mol NaCl is 1.2 M
Answer:
Q1: c. 7.2 g.
Q2: a. 0.42 M.
Explanation:
Hello there!
In this case, according to the definition of molarity as moles of solute divided by volume of solution in liters, we can proceed as follows:
Q1: Here, given the molarity and volume we can calculate the moles of the sugar as follows:
Next, since its molar mass is about 180 g/mol, the mass turns out:
Therefore, the answer is c. 7.2 g.
Q2: Here, recalling the definition of molarity, we can just plug in the 0.629 moles and 1.500 L to obtain:
Therefore, the answer is a. 0.42 M.
Best regards!
