Explanation:
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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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When PH + POH = 14
∴ POH = 14 -7 = 7
when POH = -㏒[OH-]
7 = -㏒ [OH-]
∴[OH-] = 10^-7
by using ICE table:
Mn(OH)2(s) ⇄ Mn2+ (aq) + 2OH-(aq)
initial 0 10^-7
change +X +2X
Equ X (10^-7 + 2X)
when Ksp = [Mn2+][OH-]^2
when Ksp of Mn(OH)2 = 4.6 x 10^-14
by substitution:
4.6 x 10^-14 = X*(10^-7+2X)^2 by solving this equation for X
∴ X =2.3 x 10-5 M
∴ The solubility of Mn(OH)2 in grams per liter (when the molar mass of Mn(OH)2 = 88.953 g/mol
= 2.3 x10^-5 moles/L * 88.953 g/mol
= 0.002 g/ L
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