In the laboratory you are asked to make a 0.565 m sodium bromide solution using 315 grams of water. How many grams of sodium bro
1 answer:
Answer : The mass of sodium bromide added should be, 18.3 grams.
Explanation :
Molality : It is defined as the number of moles of solute present in kilograms of solvent.
Formula used :
Solute is, NaBr and solvent is, water.
Given:
Molality of NaBr = 0.565 mol/kg
Molar mass of NaBr = 103 g/mole
Mass of water = 315 g
Now put all the given values in the above formula, we get:
Thus, the mass of sodium bromide added should be, 18.3 grams.
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Answer:
concentration of = 0.0124 = 12.4 ×10⁻³ M
concentration of = 0.0248 = 2.48 ×10⁻² M
concentration of = 0.4442 M
Explanation:
Equation for the reaction:
⇄
+
Concentration of =
= 0.469
For our ICE Table; we have:
⇄
+
Initial 0.469 0 0
Change - 2x +2x +x
Equilibrium (0.469-2x) 2x x
K =
K =
Since the value pf K is very small, only little small of reactant goes into product; so (0.469-2x)² = (0.469)²
x = 0.0124
∴ at equilibrium; concentration of = 0.0124 = 12.4 ×10⁻³ M
concentration of = 2x = 2 ( 0.0124)
= 0.0248
= 2.48 ×10⁻² M
concentration of = 0.469-2x
= 0.469-2(0.0124)
= 0.469 - 0.0248
= 0.4442 M