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An osmolarity of saline solution is 308 mosmol/L.
m(NaCl) = 9 g; the mass of sodium chloride
V(solution) = 1 L; the volume of the saline solution
n(NaCl) = 9 g ÷ 58.44 g/mol
n(NaCl) = 0.155 mol; the amount of sodium chloride
number of ions = 2
Osmotic concentration (osmolarity) is a measure of how many osmoles of particles of solute it contains per liter.
The osmolarity = n(NaCl) ÷ V(solution) × 2
The osmolarity = 0.154 mol ÷ 1 L × 2
The osmolarity = 0.154 mol/L × 1000 mmol/m × 2
The osmolarity of the saline solution = 308 mosm/L.
More about osmolarity: brainly.com/question/13258879
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The reaction is missing the Zn(s) in the reactants. The stoichiometry of the copper/zinc is 1 mole to 1 mole
Answer : The volume of 
required to neutralize is, 340 mL
Explanation :
To calculate the volume of base (NaOH), we use the equation given by neutralization reaction:
where,
are the n-factor, molarity and volume of acid which is 
are the n-factor, molarity and volume of base which is NaOH.
We are given:
Putting values in above equation, we get:
Hence, the volume of required to neutralize is, 340 mL
