Answer:
Molarity = 1.93 mol.L⁻¹
Explanation:
Molarity is the unit of concentration used to specify the amount of solute in given amount of solution. It is expressed as,
Molarity = Moles / Volume of Solution ----- (1)
Data Given;
Mass = 11.3 g
Volume = 100 mL = 0.10 L
First calculate Moles for given mass as,
Moles = Mass / M.mass
Moles = 11.3 g / 58.44 g.mol⁻¹
Moles = 0.1933 mol
Now, putting value of Moles and Volume in eq. 1,
Molarity = 0.1933 mol ÷ 0.10 L
Molarity = 1.93 mol.L⁻¹
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Answer:
2.41 g of NaCl is needed
Explanation:
Molarity of solution = (Number of moles of solute)/(Volume of solution in Litre)
Here sodium chloride (NaCl) is solute in saline solution
275 mL = 0.275 L
So, number of moles of NaCl in 275 mL saline solution =
Molar mass of NaCl = 58.44 g/mol
Number of moles = (mass)/(molar mass)
So, mass of 0.04125 moles of NaCl =
So, 2.41 g of NaCl is needed
Answer:
6.2g of NaBr are produced
Explanation:
The reaction of HBr with NaOH occurs as follows:
HBr + NaOH → NaBr + H2O
<em>Where 1 mole of each reactant produce 1 mole of NaBr</em>
To solve this question we need to find the moles of each reactant using their molar mass. With moles we can find limiting reactant and the moles (And mass) of NaBr produced, as follows:
<em>Moles HBr -Molar mass: 80.9119g/mol)-</em>
4.9g * (1mol/80.9119g) = 0.0606 moles HBr
<em>Moles NaOH -Molar mass: 40g/mol-</em>
3.86g * (1mol/40g) = 0.0965 moles NaOH
As the reaction is 1:1 and the moles of HBr < Moles NaOH, the limiting reactant is HBr and moles of NaBr produced are 0.0606 moles.
The mass of NaBr (Molar mass: 102.894g/mol) is:
0.0606 moles * (102.894g/mol) =
<h3>6.2g of NaBr are produced</h3>