Respuesta :
Answer:
The correct answer is D: 16.4 g
Explanation:
Look at the balanced equation of NaBr, It has 1:1 between Na and Br.
NaBr ---> Na+ and Br-
If the solution has 0.245 concentration of bromide ions then the solution has 0.245 concentration of sodium ions as well. Which means both have equal concentration. So the molarity of solution is 0.245 M.
Convert volume in dm3.
650ml = 650 cm3
650cm3/ 1000cm3 x 1dm3 = 0.65 dm3
Mass of a solute (g) = Molarity x Molar mass of solute x volume in dm3
= 0.245 M X 102.89 g/mol x 0.65 dm3
= 16.38 = 16.4 g
Answer:
The correct answer is D) 16.4.
The mass of sodium bromide in solution is 16.4 g
Further Explanation:
Molarity may be defined as the concentration of a solution in moles per liter. It is calculated by dividing the number of moles of a solution, n by its volume in liters, V. That is; Molarity = n/V
Therefore; when given the molarity of a solution and the volume of the solution then we can get the number of moles, by multiplying the molarity by volume; n = MV.
Additionally, the number of moles of a compound or an element can also be calculated by dividing the mass of the compound by its relative molecular mass.
In this question, we are given;
Molarity of a bromide ion = 0.245 M
But the dissociation of sodium bromide is;
NaBr(aq) → Na+(aq) + Br-(aq)
Therefore, the molarity of NaBr is 0.245 M since the mole ratio of NaBr : Br- is 1: 1.
Hence; Number of moles of NaBr = Molarity × volume
= 0.245 M × 0.650 L
= 0.15925 moles
But; number of moles = mass/ molecular mass
Molecular mass of NaBr is 102.894 g/mol
Therefore;
Mass of NaBr = Number of moles × molecular mass
= 0.15925 × 102.894 g/mol
= 16.385 g
≈ 16.4 g
Thus, the mass of sodium bromide in solution is 16.4 g
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Answer details
Grade: Middle School
Subject: Chemistry
Chapter: Mole and stoichiometry equations
Keywords: Molarity, moles, sodium bromide, molecular mass and volume.
