The concentration of calcium chloride solution required to completely react with the given concentration of **silver nitrate **is 0.06 M.
What is molarity?
Molarity is a concentration term used to represent the number of moles of solute per one liter solution. Molarity of a solution is temperature dependent.
It is given that the volume and molarity of silver nitrate solution is 1.95 L and 0.27 M respectively. The number of moles of silver nitrate is calculated as follows:
No . of moles = 0.27 M × 1.95 L
= 0.54.
As per the balanced reaction, 2 moles of silver nitrate requires only one mole of** calcium chloride**. Thus number of moles of calcium chloride required for 0.54 moles of silver nitrates is
0.54/2 = 0.27.
The **volume **of the calcium chloride solution is give 3.98 L. Thus molarity can be calculated as follows:
molarity of calcium chloride = no.of moles/volume
= 0.27/3.98 L
= 0.06 M
Therefore, the** concentration** of calcium chloride solution is 0.06 M.
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Your question is incomplete. But your complete question probably was:
The solutions of silver nitrate and calcium chloride are mixed, silver chloride precipitates out of solution according to the equation2AgNO3(aq)+CaCl2(aq)→2AgCl(s)+Ca(NO3)2(aq)a) What mass of silver chloride can be produced from 1.95 L of a 0.277 M solution of silver nitrate? Express your answer with the appropriate units b. The reaction described in Part A required 3.98 L of calcium chloride. What is the concentration of this calcium chloride solution? Express your answer with the appropriate units.
The mass of silver chloride produced from 1.95 L of 0.277 M silver nitrate is 77.42 g. The concentration of the calcium chloride solution required for the reaction is 0.0679 M.
Calculating the Mass of Silver Chloride and the Concentration of Calcium Chloride
The solutions of silver nitrate and calcium chloride are mixed, silver chloride precipitates out of solution according to the equation:
2AgNO₃(aq) + CaCl₂(aq) → 2AgCl(s) + Ca(NO₃)₂(aq)
Part A: Mass of Silver Chloride
The given 1.95 L of a 0.277 M solution of silver nitrate (AgNO₃) can be used to find the moles of AgNO₃:
Moles of AgNO₃ = Volume (L) × Concentration (M)
Moles of AgNO₃ = 1.95 L × 0.277 M = 0.54015 mol
From the balanced equation, 2 moles of AgNO₃ produce 2 moles of AgCl, so the moles of AgCl produced are also 0.54015 mol. The molar mass of AgCl is approximately 143.32 g/mol. Therefore, the mass of AgCl is:
Mass of AgCl = Moles of AgCl × Molar Mass of AgCl
Mass of AgCl = 0.54015 mol × 143.32 g/mol = 77.42 g
The mass of silver chloride produced is 77.42 g.
Part B: Concentration of Calcium Chloride
The reaction required 3.98 L of calcium chloride (CaCl₂). From the balanced equation, 1 mole of CaCl₂ reacts with 2 moles of AgNO₃, thus moles of CaCl₂ needed are:
Moles of CaCl₂ = 0.54015 mol (AgNO₃) / 2 = 0.270075 mol
The concentration of the CaCl₂ solution is determined by dividing the moles of CaCl₂ by the volume in liters:
Concentration (M) = Moles / Volume (L)
Concentration of CaCl₂ = 0.270075 mol / 3.98 L = 0.0679 M
The concentration of the calcium chloride solution is 0.0679 M.
Complete question:-
The solutions of silver nitrate and calcium chloride are mixed, silver chloride precipitates out of solution according to the equation2AgNO3(aq)+CaCl2(aq)→2AgCl(s)+Ca(NO3)2(aq)a) What mass of silver chloride can be produced from 1.95 L of a 0.277 M solution of silver nitrate? Express your answer with the appropriate units b. The reaction described in Part A required 3.98 L of calcium chloride. What is the concentration of this calcium chloride solution? Express your answer with the appropriate units.
To find the concentration of the calcium chloride solution, use the formula Molarity = Number of moles of solute / Volume of solution in liters. For example, if there are 0.27 moles of calcium chloride in 3.98 L, the molarity would be approximately 0.0677 M. Adjust the number of moles as applicable to your specific situation.
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