Compare Solution X to Solution A, Solution B, and Solution C. Beaker with a zoom view showing eight shperes, four groups of part
1 answer:
You might be interested in
Answer : The resulting concentration of ion is, 4.5 M
Explanation : Given,
Concentration of =
= 3.0 M = 3.0 mol/L
Volume of =
= 70 mL = 0.07 L
Concentration of =
= 1.0 M = 1.0 mol/L
Volume of =
= 30 mL = 0.03 L
First we have to calculate the moles of and
and,
Now we have to calculate the moles of ions.
As, 1 mole of will give 2 moles of
ions
So, 0.21 moles of will give
moles of
ions
and,
As, 1 mole of will give 1 mole of
ions
So, 0.03 moles of will give 0.03 moles of
ions
So,
Total number of moles of ions = 0.42 + 0.03 =0.45 mole
Total volume of both solution = 70 mL + 30 mL = 100 mL = 0.1 L
Now we have to calculate the concentration of ions.
Therefore, the resulting concentration of ion is, 4.5 M
Here we have to choose the right option which tells the moles of CaCl₂ will react with 6.2 moles of AgNO₃ in the reaction
2AgNO₃ + CaCl₂→ 2AgCl + Ca(NO₃)₂
6.2 moles of silver nitrate (AgNO₃) will react with B. 3.1 moles of calcium chloride (CaCl₂).
From the reaction: 2AgNO₃ + CaCl₂→ 2AgCl + Ca(NO₃)₂
Thus 2 moles of AgNO₃ reacts with 1 mole of CaCl₂
Henceforth, 6.2 moles of AgNO₃ reacts with = 3.1 moles of CaCl₂.
1 mole of CaCl₂ reacts with 2 moles of AgNO₃. Thus-
A. 2.2 moles of CaCl₂ will react with 2.2×2 = 4.4 moles of AgNO₃.
C. 6.2 moles of CaCl₂ will reacts with 6.2×2 = 12.4 moles of AgNO₃.
D. 12.4 moles of CaCl₂ will reacts with 12.4 × 2 = 24.8 moles of AgNO₃
Thus the right answer is 6.2 moles of AgNO₃ will react with 3.1 moles of CaCl₂.